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:

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

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

$90 - θ$ → angle between line and $\vec n$

θ → angle between line and plane

$\vec v$ || line = $3\hat i+5\hat j+4\hat k$

$\vec n=2\hat i+2\hat j-\hat k$

so $\vec v.\vec n=vn\cos (90-θ)$ [$v=|\vec v|,n=|\vec n|$]

$6+10-4=\sqrt{3^2+5^2+4^2}\sqrt{2^2+2^2+1^2}\sin θ$

$\sin θ=\frac{12×\sqrt{2}}{5\sqrt{2}×3×\sqrt{2}}$

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