Let $P_1: 2x+y+z+1=0$, $P_2: 2x-y+z+3=0$ and $P_3: 2x+3y+z+5=0$ be three planes, then the distance of the line of intersection of planes $P_1=0$ and $P_2=0$ from the plane $P_3=0$ is

A vector normal to $P_1=0$ is $\overrightarrow{n_1}=2\hat{i}+\hat{j}+\hat{k}$
A vector normal to $P_2=0$ is $\overrightarrow{n_2}=2\hat{i}-\hat{j}+\hat{k}$
Hence, the line of intersection is parallel to $\overrightarrow{n_1}\times \overrightarrow{n_2}$
$=\left|\begin{array}{lll}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& 1& 1\\ 2& -1& 1\end{array}\right|=\widehat{i}\mathrm{}\left(2\right)-\widehat{j}\mathrm{}\left(0\right)+\widehat{k}\mathrm{}\left(-4\right)=2\widehat{i}-4\widehat{k}$
And for point of intersection Put $z=0$ in
$P_1=0$ and $P_2=0$, i.e.
$2x+y+1=0$
$2x-y+3=0$
$4x+4=0$
$\Rightarrow x=–1$ and $y=1$
So point is $\left(-1,\mathrm{1,0}\right)$
Equation of the line of intersection is
$\frac{x+1}{1}=\frac{y-1}{0}=\frac{z-0}{-2}$
The line is parallel to $P_3=0$

Hence, distance $=\left|\frac{-2+3+5}{\sqrt{14}}\right|$
$=\frac{6}{\sqrt{14}}$