Answer:
The equation of line is:
Here,
The equation of plane is:
r→·(2i^–j^+k^)=4\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=4
Here,
Let’
\begin{array}{l}
\sin \theta=\left|\frac{(i-j+k) \cdot(2 i-j+k)}{\sqrt{1+1+1} \sqrt{4+1+1}}\right| \\
\therefore \sin \theta=\frac{2+1+1}{\sqrt{3} \sqrt{6}}=\frac{4}{\sqrt{18}}=\frac{4}{3 \sqrt{2}}=\frac{2 \sqrt{2}}{3} \\
\therefore \sin \theta=\frac{2 \sqrt{2}}{3} \\
\therefore \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)
\end{array}