$ {3}^{2} + {4}^{2} = 9 + 16 = 25 $
$ {6}^{2} = 36 $

Since, $  {3}^{2} + {4}^{2} \neq {6}^{2} , (3,4,6) $ is not a pythagorean triplet.

$ 2m, m^2+1, m^2-1$ forms are used to get Pythagorean Triplets.

To check if the set of numbers $ {a,b,c} $are Pythagorean triplets, we need to check if $ {a}^{2} + {b}^{2} = {c}^{2} $

1) For $ {8,15,17} $
$ {8}^{2} + {15}^{2} = 64 + 225 = 289 $
and $ {17}^{2} = 289 $
Hence, $ {8}^{2} + {15}^{2} = {17}^{2} $

2) For $ {16,63,65} $
$ {16}^{2} + {63}^{2} = 256 + 3969 = 4225 $
and $ {65}^{2} = 4225 $
Hence, $ {16}^{2} + {63}^{2} = {65}^{2} $

3)For $ {12,16,20} $
$ {12}^{2} + {16}^{2} = 144 + 256 = 400 $
and $ {20}^{2} = 400 $
Hence, $ {12}^{2} + {16}^{2} = {20}^{2} $

Thus, the given set of numbers are Pythagorean triplets.

Solution:

Every right triangle has side length satisfying:

one side$=10 m$

Every right triangle has side length satisfying:

Using  $ 2m, m^2+1, m^2-1$ forms

 Let us take $2m = 24$

$m=\dfrac{24}{2}=12$

if $m = 12, m^2-1=12^2= 144-1 =143$

if $ m = 12, m^2 +1=12^2= 144+1 =145$

Therefore, option A is the correct answer.

Using  $ 2m, m^2+1, m^2-1$ forms 
Let us take $2m = 18$
           

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
$7^{2}+x^{2}= (x - 2)^{2}$
49 + $x^{2}$ = $x^{2}$+ 4 -4x
Both the side $x^{2}$ will get cancelled,
45 = -4x
x = -11.25

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
$5^{2}+x^{2}= 13^{2}$
25 + $x^{2}$ = 169
$x^{2}$ = 169 - 25
$x^{2}$ = 144
Squaring on both the sides, we get
x = 12.

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
Option A: $11^{2}+60^{2}= 61^{2}$
= 121 + 3600 = 3721 is a Pythagorean triples
Option B: $16^{2}+63^{2}= 65^{2}$
= 256 + 3969 = 4225 is a Pythagorean triples
Option C: $28^{2}+45^{2}= 53^{2}$
= 784 + 2,025 = 2809 is a Pythagorean triples
Option D: $30^{2}+80^{2}= 89^{2}$
= 900 + 6400 $\neq$ 7921 is not a Pythagorean triples.

Applying the Pythagorean triples rule as $a^{2}+b^{2}= c^{2}$
Option A: $5^{2}+12^{2}= 13^{2}$
25 + 144 = 169
169 = 169 is a Pythagorean triples.

For any natural numbers m > 1, $2m, m^{2} - 1$, $m^{2} + 1$ form a Pythagorean triplet.
If we take $m^2 + 1 = 22$, then $m^2 = 21$
The value of m will not be an integer.
If we take $m^2 - 1 = 22$, then $m^2 = 23$
Again the value of m will not be an integer.
Let $2m = 22$
$m = \cfrac{22}{2}$
$m = 11$
$2m = 2 \times 11 = 22$
$m^{2} - 1$ = $11^{2} - 1$
$= 121 - 1 = 120$
$m^{2} + 1$ = $11^{2} + 1$
$= 121 + 1 = 122$
Therefore, the Pythagorean triplets are $ 22, 120, 122.$

$14, 12, 13$ is not a Pythagorean triplet.
On squaring, we get
$12^2 + 13^2\ and\ 14^2$
$144 + 169 \ and\ 196$
$313 \neq 196$

For any natural numbers m > 1, $2m, m^{2} - 1$, $m^{2} + 1$ forms a Pythagorean triplet.
If we take $m^2 + 1 = 34$, then $m^2 = 33$
The value of m will not be an integer.
If we take $m^2 - 1 = 34$, then $m^2 = 35$
Again the value of m will not be an integer.
Let $2m = 34$
$m = \dfrac{34}{2}$
$m = 17$
$2m = 2 \times 17 = 34$
$m^{2} - 1$ = $17^{2} - 1$
$= 289 - 1 = 288$
$m^{2} + 1$ = $17^{2} + 1$
$=289 + 1 = 290$
Therefore, the Pythagorean triplets are $34, 288, 290.$

$15, 114, 115$ is not a Pythagorean triplet.
On squaring, we get
$15^2 + 114^2 \ and \ 15^2$
$225 + 12996 \ and \  13225$
$13221 \neq 13225$

We can get Pythagorean triplet by using general form $2m,\ m^{2}-1,\ m^{2}+1 $
Let us first take