Any point on given line is $\left(2\mathrm{\lambda }+2,\mathrm{}\mathrm{\lambda }+\mathrm{3,5}\mathrm{\lambda }+4\mathrm{}\right)$
This must satisfy the equation of the plane
$\Rightarrow 4\mathrm{\lambda }+4-\mathrm{\lambda }-3-5\mathrm{\lambda }-4=4$
$\Rightarrow -2\mathrm{\lambda }=7\Rightarrow \mathrm{\lambda }=-\frac{7}{2}$
So, the coordinates of point are $\left(-5,-\frac{1}{2},\frac{-27}{2}\right)$
A normal vector to given plane $\left(\overrightarrow{\mathrm{n}}\right)=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$
A vector parallel to given line $\left(\overrightarrow{\mathrm{b}}\right)=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}$
A vector parallel to required line
$\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 2& -1& -1\\ 2& 1& 5\end{array}\right|=\widehat{\mathrm{i}}\left(-4\right)-\widehat{\mathrm{j}}\left(12\right)+\widehat{\mathrm{k}}\left(4\right)$
$=-4\left(\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\right)$
Equation of required line is
$\frac{\mathrm{x}+5}{1}=\frac{\mathrm{y}+\frac{1}{2}}{3}=\frac{\mathrm{z}+\frac{27}{2}}{-1}$