The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$  is the hypotenuse of an  isosceles right-angled triangle whose opposite vertex is (7 ,2,4). Then which of the following is not the side of the triangle?

Given are vertex A(7 ,2, 4) and line

$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$

$\text{ General point on above line }B\equiv (5\lambda -6,3\lambda -10,8\lambda -14)$

$\text{ Direction ratios of line }AB\text{ are }<5\lambda -13,3\lambda -12,8\lambda -18>$

$\text{ Direction ratios of line }BC\text{ are }<5,3,8>$

$\text{ Since angle between }AB\text{ and }BC\text{ is }\pi /4\text{, we have }$

$\cos \frac{\pi }{4}=\frac{(5\lambda -3)5+3(3\lambda -12)+8(8\lambda -18)}{\sqrt{5^2+3^2+8^2}\cdot \sqrt{(5\lambda -13{)}^2+(3\lambda -12{)}^2+(8\lambda -18{)}^2}}$

$\text{ Squaring and solving, we get }\lambda =3,2$

$\text{ Hence, equation of lines are }\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}\text{ and }\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$