Question

If equation of the plane through the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$ is $ax-by+cz+4=0$, then the value of $a^2+b^2+c$ is

Answer 1

Given: $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$
Let normals of plane is $\ell ,m,n$ is perpandicular to DR's of line $(2,-3,5)$ and DR's of plane $(1,-1,1)$
Let equation of a plane containing the line be $\ell (x-1)+m(y+2)+nz=0\cdots \left(i\right)$ then
$2\ell -3m+5n=0\cdots \left(ii\right)$ and
$\ell -m+n=0\cdots \left(iii\right)$
Solving $\left(ii\right),\left(iii\right)$
$\frac{\ell }{2}=\frac{m}{3}=\frac{n}{1}=k$
putting in equation $\left(i\right)$
$\Rightarrow 2(x-1)+3(y+2)+z=0$
$\Rightarrow 2x+3y+z+4=0$
On comparing with $ax-by+cz+4=0$
We get $a=2,b=-3,c=1$
$a^2+b^2+c=4+9+1=14$