Question

Let the equation of the plane passing through the line $x-2 y-z-5=0=x+y+3 z-5$ and parallel to the line $x+y+2 z-7=0=2 x+3 y+z-2$ be $a x+b y+c z=65$. Then the distance of the point $(a, b, c)$ from the plane $2 x+2 y-z+16=0$ is _______

Answer 1

Equation of plane is
$ ( x -2 y - z -5)+ b ( x + y +3 z -5)=0 $
$ \begin{vmatrix} 1+ b & -2+ b & -1+3 b \\1 & 1 & 2 \\2 & 3 & 1\end{vmatrix}=0$
$\Rightarrow b =12 $
$ \therefore \text { plane is } 13 x +10 y +35 z =65$
Distance from given point to plane $=9$