Angle between the line $\frac{x-1}{3}=\frac{y+1}{5}=\frac{z-2}{4}$ and the plane $2 x+2 y-z=3$ is:

$\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$ $\sin ^{-1}\left(\sqrt{\frac{2}{5}}\right)$ $\sin ^{-1}\left(\frac{\sqrt{2}}{5}\right)$ $\sin ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$

$\sin ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$

The correct answer is Option (4) - $\sin ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$