Question

If the distance of the plane passing through the point $P(1,1,1)$ and perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ from the origin is $\frac{k}{5}$ then the value of $k$ is

Answer 1

Equation of the plane passing through $(1,1,1)$ is given by, $A(x-1)+B(y-1)+C(z-1)=0\dots \left(1\right)$

Since the line is perpendicular to the plane $\left(1\right)$

$\therefore 3(x-1)+0(y-1)+4(z-1)=03x+4z-7=0$

Hence, distance from $(0,0,0)$ to the plane $\left(1\right)$ is,

$d=\left|\frac{-7}{\sqrt{3^2+4^2}}\right|=\frac{7}{5}$
$\therefore k=7$