The foot of the perpendicular drawn from the origin to the plane is $(4,-2,-5)$. Hence, the equation of the…
The foot of the perpendicular drawn from the origin to the plane is $(4,-2,-5)$. Hence, the equation of the plane is
$4 x-2 y+5 z=-5$
$4 x-2 y-5 z=45$
$4 x+2 y-5 z=37$
$4 x+2 y+5 z+13=0$
Solution
D.r's of normal to the plane $\langle 4-0,-2-0,-5-0\rangle \equiv\langle 4,-2,-5\rangle$
Hence equation of the plane $4 x-2 y-5 z=\lambda$
But it passes through $(4,-2,-5)$
$\begin{aligned}
& \Rightarrow 4 \times 4-2 \times(-2)-5 \times(-5)=\lambda \\
& \Rightarrow \lambda=45
\end{aligned}$