The angle between the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{x+5}{6}$ and the plane $2x+10y-11z=5$ is:

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

Line = $\frac{x+2}{3}=\frac{y-3}{2}=\frac{x+5}{6}$, Plane = $2x+10y-11z=5$

vector || to line = $3\hat i+2\hat j+6\hat k=(\vec{v_1})$

vector ⊥ plane = $2\hat i+10\hat j-11\hat k=(\vec{v_2})$

$|\vec{v_1}.\vec{v_2}|=|\vec{v_1}||\vec{v_2}|\cos θ$   (θ → angle between normal to plane and line)

if $α$ is angle between plane and line

$⇒|6+20-66|=\sqrt{3^2+2^2+6^2}\sqrt{2^2+10^2+11^2}\cos θ⇒α=90°-θ$

$\cos θ=\frac{40}{7×15}$

so $\cos(90-α)=\frac{8}{21}$

$α=\sin^{-1}(\frac{8}{21})$