$P(1,2,3),$ Plane :$x-y+z=5$

Let the given points be A($1,2,3$) and B($3,2,-1$) and let the point equidistant from A and B be P($x,y,z$) 

Let the given points be $A(1,1,1)$

$P(2,2,1),Q(5,1,-2)$

$\begin{array}{l} x=3\, \, \, ,y=7+t\, \, ,z=1+t \ A\left( { 3,7+t,1+t } \right)  \ 0.\left( { 3-5 } \right) +1\left( { 7+t-1 } \right) +1\left( { 1+t-3 } \right) =0 \ 6+t+t-2=0 \ 2t=-4 \ t=-2 \ A:\left( { 3,5,-1 } \right)  \ dis\tan  ce=\sqrt { 4+16+16 }  \ =6 \ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

Planes are  $2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$

Origin is $O(0,0,0)$ and given point is $A(4,5,6)$

Let the extremities of the diagonal of a square be $(1,-2,3)$ and $B(2,-3,5)$.
Then $AB$ is given by $ {({1}^{2} + {1}^{2} + {2}^{2})}^{0.5} $ = $ \sqrt{6} $
Hence, length of the side $ = \sqrt {3} $
So, area of square will be $ \sqrt{3} \times \sqrt{3}  = 3$

Equation of sphere passes through origin is given by

Consider the problem,

Consider the problem,

$\begin{array}{l} According\, to\, question: \ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \ Now, \ line\, of\, equ: \ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \ Now\, find\, x,y\, & \, z\, \, coordinates: \ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \ \Rightarrow \frac { { y-1 } }{ 4 } =13 \ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \,  \ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$

If $({ l } _{ 1 }{ m } _{ 1 }{ n } _{ 1 })$ and  $({ l } _{ 2 }{ m } _{ 2 }{ n } _{ 2 })$ are directions of two $\bot$ lines then,

If a point (x,y) is equidistant from A(a,b) and B(c,d) then,

Let the vertices of triangle are 

${x}^{2}+{y}^{2}=0$

The point is $P(1,2,3)$ so distance of point from $y$ axis is
${=}$ $\sqrt{{(1)}^{2} + {(3)}^{2} } $  

Let $(x, y, z)$ be the point.

If $a$ is the length of a side of square then the length of diagonal is given by $\sqrt{2}a$. Distance between two given  points is $\sqrt{(1-4)^2+(-2-2)^2+(3-3)^2}=5=\sqrt{2}a$. Hence the area is given by $a^2=\dfrac{25}{2}$.

Center of the circum circle of a right angle triangle is on the mid of the hypotenuse. 

Distance=$\sqrt{(x _2-x _1)^2+(y _2-y _1)^2)}$

Let $R(x,0,0)$ be the point on $x$-axis which is equidistant from $P(4,3,1)$ and $Q(-2,-6,-2)$

Given points are $A(2,-3,-1)$ and $B(8,-1,2)$
Therefore direction  ratio of $AB$ are $l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$ or $l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$
Hence, coordinates of a point $14$ unit from point $A$ is given as,$(2+14l,-3+14m,-1+14n)$
$\Rightarrow (14,1,5)$ or $(-10,-7,-7)$
Hence, option 'A' is correct.

Given $C _1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$
and $C _2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$
So centre and radius of $C _1$ and $C _2$ are $(10,0)$, $(-15,0)$ and $6,9$ respectively.
Then, distance between $C _1$ and $C _2$ is $\sqrt{(15+10)^{2}+(0-0)^{2}}=25$. 

Let the coordinates of point $Q$ be $(a,b,c)$

Equation of the line through $\left( 1,-2,3 \right) $ parallel to the line $\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 } $ is

$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$ (say)   ...$(1)$

Then any point on $(1)$ is $\left( 2r+1,3r-2,-6r+3 \right) $.

If this point lies on the plane $x-y+z=5$, then 

$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 } $

Hence, the point is $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $

Distance between $\left( 1,-2,3 \right) $ and $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $

$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 }  \right)  } =\sqrt { \dfrac { 49 }{ 49 }  } =1$

Let $A=(4,-5,1)$
$B=(3,-4,0)$
$C=(6,-7,3)$
$D=(7,-8,4)$
Now let the quadrilateral be $ABCD$.
Then
$AB=-i+j-k$ $|AB|=\sqrt{3}$
$BC=3i-3j+3k$ $|BC|=3\sqrt{3}$
$CD=i-j+k$  $|CD|=\sqrt{3}$
$AD=3i-3j+3k$ $|AD|=3\sqrt{3}$.
Hence opposite sides are equal and parallel.
Therefore the above points form a parallelogram.
Now all the sides are not equal.
Hence it cannot qualify as a rhombus or square.
Now dot product of $AB$ and $BC$ is not zero.
Hence adjacent sides are not perpendicular to each other.
Therefore it is not a rectangle also.

Let $P$ be the required point $\displaystyle \left ( x,y,z \right )$ and the point
$A, B, C$ and $O$ are $\displaystyle \left ( a,0,0 \right ),\left ( 0,b,0 \right ),\left ( 0,0,c \right )$ and $\displaystyle \left ( 0,0,0 \right )$ ;

 Vertices of triangle are $(0,4,0),(3,4,0)$ and $(0,4,4)$

(i) Distance between $4i + 3j - 6k$ and $-2i + j - k$ is given by $\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(2,3,-1)$, $(2,6,2)$
Distance $=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } } $
                $=\sqrt { 18 } $
                $=3\sqrt { 2 }$

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(-1,1,3)$, $(0,5,6)$
Distance $=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } $
                $=\sqrt { 26 } $

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $-2\hat i+3\hat j+5\hat k$, $7\hat i-\hat k$
Distance $=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } } $
                $=\sqrt { 126 } $
                $=3\sqrt { 14 } $

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $4\hat i+3\hat j-6\hat k$, $-2\hat i+\hat j-\hat k$
Distance $=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } } $
                $=\sqrt { 65 } $

Let $A=(3,-5,1), B = (-1,0,8)$ and $C=(7,-10,-6)$

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by 

($ \dfrac{b}{3}, 1 , -2 $)

Hence, $ \dfrac{b}{3} $ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

Let the points $A(2,5,1) B(1,4,-3)$ and $C(-2,7,-3)$
Now using distance formula in $3D$, we have
$AB {=}$$\sqrt{18}$, $BC {=}$$\sqrt{18}$, and $AC {=}$$\sqrt{36}$
Since, ${AB}^{2}+{BC}^{2}$${=}$${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $AC$.
Therefore by using section formula (1:1), circumcentre ${=}(0,6,-1)$

The $4$ points are as given.
We calculate their individual distance from the origin.
$OP =$ $ {({5}^{2} + {6}^{2})}^{0.5} $ = $ {(61)}^{0.5} $
$OQ =$ $ {({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5} $ = $ {(66)}^{0.5} $
$OR =$ $ {({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5} $ = $ {(62)}^{0.5} $
$OS =$ $ {({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5} $= $ {(290)}^{0.5} $
Hence, $P$ is the nearest to the origin.

Keeping the above points as vertices and using the distance formula, 
$D=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}+(z _{2}-z _{1})^{2}}$
We get that the sides of the parallelogram formed by the above lines are equal and all the sides are of $7$ units.
Hence the parallelogram will be either a rhombus or a square. For the parallelogram to be square, all the adjacent sides of the parallelogram should make an angle of $\dfrac{\pi}{2}$.
Consider the vector equation of $AB$ as $(5-(-1))i'+(-1-(-3))j'+(1-4)k'$
$6i'+2j'-3k'$
Consider the vector equation of $BC$ as $(7-5)i'+(-4-(-1))j'+(7-1)k'$
$2i'-3j'+6k'$
Taking dot product, we get $6(2)+2(-3)-3(6)$
$=12-6-18$
$=-12$
Thus the parallelogram is a rhombus.

Given that,

Dimensions of the hall x $=24cm\times 8cm\times 6cm$

Now Leght of the longest pole which can be accommodated in the hall x $=\sqrt{{{24}^{2}}+{{8}^{2}}+{{6}^{2}}}=26cm$

We know the distance formula:

$d=\sqrt{(x _2-x _1)^2+(y _2-y _1)^2+(z _2-z _1)^2}$

The coordinates are $(-3, 6, 7)$ and$ (2, -1, 4)$

$d=\sqrt{(2-(-3))^2+((-1)-6)^2+(4-7)^2}$

$d=\sqrt{(5)^2+((-7)^2+(-3)^2}$

$d=\sqrt{25+49+9}$

$d=\sqrt{83}$

$d = 9.11$

$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$
$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$
$\lambda^2+16\lambda^2+8\lambda^2=36$
$\lambda=\pm \frac {6}{5}$
$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$

$x+y+2z-3=0=2x+3y+4z-4$ at $z-axis, x=y=0$

Speed of the man in still water $\cfrac{2}{15/60}$ = $8$ km/hr
Speed of the man in downstream = $\cfrac{4}{20/60}$ = $12$ km/hr
Speed of stream = $4$ km/hr
Distance covered by paper boat in $2$ $\frac{1}{2}$ hours = $\cfrac{5}{2} \, \times$  $4$ = $10$ km 

$\begin{matrix} then\, a=? \ The\, equation\, lineAB\, is\,  \ \Rightarrow \dfrac { { x-8 } }{ { 5-8 } } =\dfrac { { y+7 } }{ { 2+7 } } =\dfrac { { z-a } }{ { 4-a } } ....\left( 1 \right)  \ po{ int }\, c\, lies\, in\, the\, line\, AB \ po{ int }\left( { 6,-1,2 } \right) satisfy\, eq\left( 1 \right)  \ \Rightarrow \dfrac { { -2 } }{ { -3 } } =\dfrac { { -1+7 } }{ 9 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow \dfrac { 2 }{ 3 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow 8-2a=6-3a \ \Rightarrow 3a-2a=6-8 \ a=-2 \  \end{matrix}$

Eliminating t from the given equation, we get the equation of the path $\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $

AB${=}$ $\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$
AB${=}$ $\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB${=}$ $\sqrt{49}$
AB${=}$ 7
Similarly you find that BC${=}$ $\sqrt{49}$  CD${=}$ 7  and DA${=}$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC${=}$ $\sqrt{122}$
similarly BD${=}$ $\sqrt{74}$ 
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through  BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus

AB${=}$ $\sqrt{{(-1-1)}^{2}+{(-2-2)}^{2}+{(-1-3)}^{2}}$
AB${=}$ $\sqrt{{(-2)}^{2}+{(-4)}^{2}+{(-4)}^{2}}$
AB${=}$ $\sqrt{36}$
AB${=}$ 6
Similarly you find that BC${=}$ $\sqrt{43}$  CD${=}$ 6  and DA${=}$ $\sqrt{43}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$
AC${=}$ $\sqrt{3}$
similarly BD${=}$ $\sqrt{155}$ 
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through  CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram

Given points are, $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=

\left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle

\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}

\right )$

Length $AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$
Length $BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$
Length $CD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$
Length $AD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$

Angle between two vectors $\overline{AB} = p _{1}\hat{i}+q _{1}\hat{j}+r _{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$ and $\overline{BC} = p _{2}\hat{i}+q _{2}\hat{j}+r _{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$ is $cos\theta = 0 \Rightarrow \theta = 90^o$

All sides are equal and angle between $AB$ and $BC$ is $90^o$, Hence, its a square.

AB${=}$ $\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$
AB${=}$ $\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$
AB${=}$ $\sqrt{11}$
Similarly you find that BC${=}$ $\sqrt{6}$  CD${=}$ $\sqrt{11}$  and DA${=}$$\sqrt{6}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$
AC${=}$ $\sqrt{17}$
similarly BD${=}$ $\sqrt{17}$ 
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through  BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle

The plane forming the parallelipiped are

Let $P(x,y,z)$ be the required point.

The distance between the points $(1,0,5)$ and $(-3,6,3)$ is given below:

$\dfrac{|\begin{vmatrix}2-1&4-2  &5-3 \2 & 3 &4 \ 3& 2 & 5\end{vmatrix}|}{|\begin{vmatrix}i&j  &k \2 & 3 & 4\3 & 2 & 5\end{vmatrix}|}=\dfrac{1}{\sqrt{78}}$