<s>[INST] 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 $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.

My Attempt:
Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.
Let the equation of  $AB$ be $\dfrac{x-7}{a_1}=\dfrac{y-2}{b_1}=\dfrac{z-4}{c_1}$
and the equation of $BC$ be $\dfrac{x-7}{a_2}=\dfrac{y-2}{b_2}=\dfrac{z-4}{c_2}$
As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$
Also as the triangle is a isosceles right triangle, so angle between $AC$ and 
$BC$ will be same as the angle between $CA$ and $BA$.  Let it be $\theta$.
So $\cos\theta=\dfrac{5a_1+3b_1+8c_1}{\sqrt{5^2+3^2+8^2}\sqrt{a_1^2+b_1^2+c_1^2}}=\dfrac{5a_2+3b_2+8c_2}{\sqrt{5^2+3^2+8^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Squaring both sides we get
$\dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=\dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$
But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.
 [/INST] Sides AB and AC make an angle of 45∘ with the given line.
Let B=(−6+5t,−10+3t,−14+8t)
AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^
cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√
Squaring and simplifying t2–5t+6=0⟹t=2or3
Taking 2 for B and 3 for C.
AB→=−3i^−6j^−2k^
AC→=2i^−3j^+6k^
Equation of AB is x−73=y−26=z−42
Equation of AC is  x−72=y−2−3=z−46
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