In the Cartesian coordinate system, the first quadrant is located in the upper right, the second quadrant is located in the upper left, the third quadrant is located in the lower left, and the fourth quadrant is located in the lower right. Therefore, the point (3, -4) is located in the fourth quadrant.

The equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Given that the slope is -3 and the line passes through the point (2, 5), we can substitute these values into the equation to find the y-intercept: 5 = -3(2) + b, which gives b = 11. Therefore, the equation of the line is y = -3x + 11.

The distance between two points (x1, y1) and (x2, y2) is given by the formula: distance = √((x2 - x1)^2 + (y2 - y1)^2). Substituting the values of the given points, we get: distance = √((-2 - 4)^2 + (7 - 3)^2) = √((-6)^2 + (4)^2) = √(36 + 16) = √52 = 5√5.

The angle between two lines can be found using the formula: angle = arctan(|m2 - m1| / (1 + m1 * m2)), where m1 and m2 are the slopes of the lines. Substituting the slopes of the given lines, we get: angle = arctan(|2 - (-1)| / (1 + 2 * (-1))) = arctan(3 / (-1)) = arctan(-3) ≈ 45°. Therefore, the angle between the lines is approximately 45°.

The standard equation of a circle with center at the origin and radius r is given by x^2 + y^2 = r^2. Therefore, the equation x^2 + y^2 = 9 represents a circle with center at the origin and radius 3.

The standard equation of a parabola with vertex at the origin and focus at (0, p) is given by y^2 = 4px. Substituting p = 2, we get the equation y^2 = 4x.

The standard equation of a hyperbola with center at the origin, vertices at (±a, 0), and foci at (±c, 0) is given by x^2 / a^2 - y^2 / b^2 = 1, where c^2 = a^2 + b^2. Substituting a = 3 and c = 5, we get: 5^2 = 3^2 + b^2, which gives b^2 = 16. Therefore, the equation of the hyperbola is x^2 / 9 - y^2 / 16 = 1, or equivalently, x^2 - y^2 = 25.

The slope of the tangent line to a circle at a given point is equal to the negative reciprocal of the slope of the radius vector from the center of the circle to that point. The radius vector from the center (0, 0) to the point (3, 4) has a slope of 4/3. Therefore, the slope of the tangent line is -3/4. Substituting this slope and the point (3, 4) into the point-slope form of a line, we get: y - 4 = (-3/4)(x - 3), which simplifies to y = -4x + 7.

To find the area of the triangle, we can first find the points of intersection of the three lines. The intersection of y = 2x + 1 and y = x - 1 is (2, 3). The intersection of y = 2x + 1 and x = 3 is (3, 7). The intersection of y = x - 1 and x = 3 is (3, 2). The area of the triangle can then be calculated using the formula: area = 0.5 * |(x1 * y2 + x2 * y3 + x3 * y1) - (y1 * x2 + y2 * x3 + y3 * x1)|, where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the three vertices of the triangle. Substituting the values of the vertices, we get: area = 0.5 * |(2 * 3 + 3 * 2 + 3 * 3) - (3 * 2 + 7 * 3 + 2 * 3)| = 0.5 * |(6 + 6 + 9) - (6 + 21 + 6)| = 0.5 * |21 - 33| = 0.5 * 12 = 6. Therefore, the area of the triangle is 6 square units.

To find the equation of the plane, we can use the vector equation of a plane: r = r0 + s * v1 + t * v2, where r0 is a point on the plane, v1 and v2 are vectors parallel to the plane, and s and t are scalar parameters. We can choose r0 to be the point (1, 2, 3), and v1 and v2 to be the vectors (2, 3, 4) - (1, 2, 3) = (1, 1, 1) and (3, 4, 5) - (1, 2, 3) = (2, 2, 2), respectively. Substituting these values into the vector equation, we get: r = (1, 2, 3) + s * (1, 1, 1) + t * (2, 2, 2). To convert this into an equation of the plane, we can set s = x - 1 and t = y - 2, which gives: r = (1, 2, 3) + (x - 1) * (1, 1, 1) + (y - 2) * (2, 2, 2). Expanding this equation, we get: r = (1 + x - 1, 2 + y - 2, 3 + x - 1 + 2y - 4) = (x, y, x + 2y - 3). Therefore, the equation of the plane is x + 2y + 3z = 14.

The distance from a point (x0, y0, z0) to a plane Ax + By + Cz + D = 0 is given by the formula: distance = |Ax0 + By0 + Cz0 + D| / √(A^2 + B^2 + C^2). Substituting the values of the point and the plane, we get: distance = |2 * 2 + 3 * 3 - 4 * 4 + 10| / √(2^2 + 3^2 + (-1)^2) = |4 + 9 - 16 + 10| / √(4 + 9 + 1) = |7| / √14 = 7 / √14 ≈ 2. Therefore, the distance from the point (2, 3, 4) to the plane 2x + 3y - z = 10 is approximately 2 units.

To find the line of intersection of two planes, we can solve the system of equations formed by the two plane equations. Substituting the second equation into the first equation, we get: x + y + (2x - y + z) = 6, which simplifies to 3x + z = 4. This equation represents a plane parallel to the y-axis. To find the line of intersection, we can choose a point on this plane, such as (1, 0, 1), and find the direction vector of the line. The direction vector can be found by taking the cross product of the normal vectors of the two planes. The normal vector of the first plane is (1, 1, 1), and the normal vector of the second plane is (2, -1, 1). Taking the cross product of these vectors, we get: (1, 1, 1) x (2, -1, 1) = (-2, -3, 3). Therefore, the direction vector of the line of intersection is (-2, -3, 3). Using the point (1, 0, 1) and the direction vector (-2, -3, 3), we can write the parametric equations of the line of intersection as: x = 1 - 2t, y = 0 - 3t = -3t, z = 1 + 3t.

The angle between two planes can be found using the formula: angle = cos^-1((n1 · n2) / (|n1| * |n2|)), where n1 and n2 are the normal vectors of the planes. The normal vector of the first plane is (2, 1, -1), and the normal vector of the second plane is (1, -1, 2). Substituting these values into the formula, we get: angle = cos^-1(((2, 1, -1) · (1, -1, 2)) / (|(2, 1, -1)| * |(1, -1, 2)|)) = cos^-1((2 - 1 + 2) / (√(2^2 + 1^2 + (-1)^2) * √(1^2 + (-1)^2 + 2^2))) = cos^-1(3 / √6 * √6) = cos^-1(3 / 6) = cos^-1(1/2) ≈ 60°. Therefore, the angle between the planes 2x + y - z = 3 and x - y + 2z = 5 is approximately 60°.

The standard equation of a sphere with center at (h, k, l) and radius r is given by (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2. Substituting the values of the center and radius, we get: (x - 2)^2 + (y + 1)^2 + (z - 3)^2 = 4^2 = 16. Therefore, the equation of the sphere with center at (2, -1, 3) and radius 4 is (x - 2)^2 + (y + 1)^2 + (z - 3)^2 = 16.