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). Find the equation of the remaining sides.

 $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$
General point on above line.
$B\equiv (5\lambda –6,3\lambda –10,8\lambda –14)$ 
Dr’s of line  $AB\propto (5\lambda –6–7),(3\lambda –10–2),(8\lambda –14–4)$
  $\propto 5\lambda –13,3\lambda –12,8\lambda –18$
Dr’s of line  $BC\propto 5,3,8$
Since angle between AB & BC is $\frac{\pi }{4}$  
 $$\begin{gathered} \cos \frac{\pi }{4}=\frac{(5\lambda –3)5+3(3\lambda –12)+8(8\lambda –18)}{\sqrt{5^2+3^2+8^2}.\sqrt{{(5\lambda –13)}^2+{(3\lambda –12)}^2+{(8\lambda –18)}^2}} \\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{25\lambda +9\lambda +64\lambda –15–36–144}{\sqrt{98}\sqrt{{(5\lambda –13)}^2+{(3\lambda –12)}^2+{(8\lambda –18)}^2}} \end{gathered}$$
 
After squaring & solving above equation we gert
 $\lambda =3,2$
Hence equation of line are
 $$\begin{gathered} \frac{x–7}{2}=\frac{y–2}{–3}=\frac{z–4}{6} \\ \&\frac{x–7}{–3}=\frac{y–2}{–6}=\frac{z–4}{–2}\Rightarrow \frac{x–7}{3}=\frac{y–2}{6}=\frac{z–4}{2} \end{gathered}$$