Question

Let the equation of the plane which contains the line $x=\frac{y-3}{2}=\frac{z-5}{3},$ and which is perpendicular to the plane $2x+7y-3z=1.$ be $kx-my-z+p=0.$ Find $p-k-m$ ?

Answer 1

Plane through given line is
$Ax+B(y-3)+C(z-5)=0.$ ... $\left(1\right)$
where $A.1+B.2+C.3=0.$ ... $\left(2\right)$
Also the plane is perpendicular to $2x+7y-3z=1$
$\therefore 2A+7B+C(-3)=0$ ... $\left(3\right)$
Solving $\left(2\right)$ and $\left(3\right),$ we get
$\frac{A}{-27}=\frac{B}{9}=\frac{C}{3}$ or $\frac{A}{9}=\frac{B}{-3}=\frac{C}{-1}.$
Hence, the required plane is
$9x-3(y-3)-1(z-5)=0$
$\Rightarrow 9x-3y-z+14=0.$
$\Rightarrow p-k-m=2$