Let $\theta$ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have
$\vec{r} =\left(-\hat{i} + 3\hat{k}\right) + \lambda \left(2\hat{i} + 3\hat{j} + 6\hat{k}\right) $ and
$\vec{r} .\left(10 \hat{i} + 2 \hat{j} -11\hat{k}\right) = 3$
Here, $ \vec{b} = 2\hat{i} + 3\hat{j} +6\hat{k} $ and $ \vec{n} = 10 \hat{i} + 2 \hat{j} - 11 \hat{k} $
$ \sin\phi = \left|\frac{\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right) .\left(10 \hat{i} + 2\hat{j} -11\hat{k}\right)}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+11^{2}}}\right|$
$ = \left|\frac{-40}{7 \times15}\right| = \left|\frac{-8}{21}\right|= \frac{8}{21}$ or $ \phi =\sin^{-1} \left(\frac{8}{21}\right) $