Find the equation of the plane if the foot of the perpendicular from origin to the plane is (1, 3, –5)

Let origin 'O' = (0, 0, 0) Let foot of perpendicular be M = (1, 3, –5)
The d.rs of the plane = d.r's of $\overline{\mathrm{OM}}$
(a, b, c) = (1–0, 3–0,–5–0)

 $\therefore$(a, b,c) = (1, 3, –5)
The equation of required plane through (1, 3, –5) is
$\begin{array}{l}\mathrm{a}\left(\mathrm{x}-{\mathrm{x}}_1\right)+\mathrm{b}\left(\mathrm{y}-{\mathrm{y}}_1\right)+\mathrm{c}\left(\mathrm{z}-{\mathrm{z}}_1\right)=0\\ 1(\mathrm{x}-1)+3(\mathrm{y}-3)-5(\mathrm{z}+5)=0\\ \mathrm{x}-1+3\mathrm{y}-9-5\mathrm{z}-25=0\\ \Rightarrow \mathrm{x}+3\mathrm{y}-5\mathrm{z}-35=0\end{array}$