The equation of the given line is $\frac{x-2}{2}$ = $\frac{y+1}{4}$ = $\frac{z-2}{2}$ ... (1) Any point on the given line is (3λ + 2, 4λ - 1 , 2λ + 2) If this point lies on the given plane x – y + z – 5 = 0, then 3λ + 2 -(4λ - 1) + 2λ + 2 - 5 = 0 ⇒ λ = 0 Putting λ = 0 in (3λ + 2, 4λ - 1 , 2λ + 2) , we get the point of intersection of the given line and the plane is (2, -1, 2). Let θ be the angle between the given line and the plane ∴ sin θ = $\frac{\vec{a} \cdot \vec{b}}{\left|\vec{a}\right|\left|\vec{b}\right|}$ = $\frac{(\widehat{3i} + \widehat{4j} + \widehat{2k}) \cdot (\hat{i} - \hat{j} + \hat{k})}{\sqrt{3^2+4^2+2^2} \sqrt{1^2+1^2+1^2}}$ = $\frac{3 - 4 + 2}{\sqrt{29} \sqrt{3}}$ = $\frac{1}{\sqrt{87}}$ ⇒ θ = $\sin^{-1}\left(\frac{1}{\sqrt{87}}\right)$ Thus, the angle between the given line and the given plane is $\sin^{-1}\left(\frac{1}{\sqrt{87}}\right)$ OR The equation of the given planes are $\vec{r} \cdot \hat{i} + \widehat{2j} + \widehat{3k} - 4$ = 0 ... (1) $\vec{r} \cdot \widehat{2i} + \hat{j} - \hat{k} + 5$ = 0 ... (2) The equation of the plane passing through the intersection of the planes (1) and (2) is $\left|\vec{r} \cdot \hat{i} + \widehat{2j} + \widehat{3k} - 4\right|$ + λ $\left|\vec{r} \cdot \widehat{2i} + \hat{j} - \hat{k} + 5\right|$ = 4 - 5λ ... (3) Given that plane (3) is perpendicular to the plane $\vec{r} \cdot \widehat{5i} + \widehat{3j} - \widehat{6k} + 8$ = 0 1 + 2λ × 5 + 2 + λ × 3 + 3 - λ × - 6 = 0 ⇒ 19λ - 7 = 0 ⇒ λ = $\frac{7}{19}$ Putting λ = $\frac{7}{19}$ in (3), we get = 4 - $\frac{35}{19}$ ⇒ $\vec{r} \cdot \left(\frac{33}{19}\hat{i} + \frac{45}{19}\hat{j} + \frac{50}{19}\hat{k}\right)$ = $\frac{41}{19}$ ⇒ $\vec{r} \cdot \widehat{33i} + \widehat{45j} + \widehat{50k}$ = 41. This is the equation of the required plane.