JEE Main 2018 — Three Dimensional Geometry Question with Solution

Normal to $3x+4y+z=1$ is $3\hat{i}+4\hat{j}+k$. Normal to $5x+8y+2z=-14$ is $5\hat{i}+8\hat{j}+2\hat{k}$ The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors. Therefore the direction ratios of the line is $-\hat{j}+4\hat{k}$ Hence the angle between the plane $x+y+z+5=0$ and the intersection line is ${\sin }^{-1}\left(\frac{-1+4}{\sqrt{17}\sqrt{3}}\right)={\sin }^{-1}\left(\sqrt{\frac{3}{17}}\right)$