The angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y -11z=3$ is :

$sin^{-1}(\frac{8}{21})$

The correct answer is Option (4) → $\sin^{-1}(\frac{8}{21})$

$\vec v$(parallel to line) = $2\hat i+3\hat j+6\hat k$

$\vec n$ (perpendicular to plane) = $10\hat i+2\hat j-11\hat k$

θ → angle between line and plane

$90 - θ$ → angle between normal and plane

$|\vec v.\vec n|=|\vec v||\vec n|\cos(90 - θ)$

$=|-40|=7×15\sin θ⇒\sin θ=\frac{40}{7×15}$

$θ=\sin^{-1}(\frac{8}{21})$