By using equations of the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10 x+2 y-11 z-3=0$, answer the following questions.

The acute angle between the line and the plane is

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

$\vec{n}$ of plane = $10\hat{i} + 2\hat{j} - 11\hat{k}$

$|\vec{n}| = n = \sqrt{10^2+2^2+11^2} = 15$

vector $\vec{v}$ || line = $2\hat{i} + 3\hat{j} + 6\hat{k}$

$|\vec{v}| = v = \sqrt{2^2+3^2+6^2} = 7$

θ → angle between line and place

90 - θ → angle between normal and line

$\vec{n} . \vec{v} = nv \cos (90 - θ) ⇒ \frac{|20+6-6|}{15 \times 7}=\sin \theta$

so  $\sin \theta=\frac{8}{21} \rightarrow \theta= \sin^{-1}(\frac{8}{21})$

Option: B