The coordinates of the foot of perpendicular drawn from origin to the plane
$2x-y+5z-3=0$ are ________

Use foot of perpendicula \((x, y, z)\) of a point \(\left(x_{1}, y_{1}, z_{1}\right)\) in a plane \(a x+b y+c z\) \(+d=0\) is given by \(\frac{x-x_{1}}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{c}}=\underline{-(\mathrm{ax}}\)
Given equation of plane is \(2 x-y+5 z-3\) \(=0\)
\(\therefore\) Foot of perpendicular drawn from origin to the given plane
\(
\begin{aligned}
&\Rightarrow \frac{x-0}{2}=\frac{y-0}{-1}=\frac{z-0}{5} \\
&=\frac{-(0+0+0-3)}{(2)^{2}+(-1)^{2}+(5)^{2}} \\
&\Rightarrow \frac{x}{2}=\frac{y}{-1}=\frac{z}{5}=\frac{3}{30}=\frac{1}{10} \\
&\therefore \frac{x}{2}=\frac{1}{10}, \frac{y}{-1}=\frac{1}{10}, \frac{z}{5}=\frac{1}{10} \\
&\Rightarrow x=\frac{1}{5}, y=-\frac{1}{10}, z=\frac{1}{2}
\end{aligned}
\)