An angle between the plane $x+y+z=5$ and the line of intersection of the planes, $3x+4y+z-1=0$ and $5x+8y+2z+14=0$ is

Let $\overrightarrow{n}$ be the vector along the line of intersection of given planes.

$\Rightarrow \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 1\\ 5& 8& 2\end{array}\right|=\hat{i}\left(0\right)-\hat{j}(6-5)+\hat{k}(24-20)=-\hat{j}+4\hat{k}$

Now, angle between the plane $x+y+z=5$ and the line of intersection of the planes $3x+4y+z-1=0$ and $5x+8y+2z+14=0$ is equal to $\frac{\pi }{2}-{\cos }^{-1}\left(\frac{-1+4}{\sqrt{3 } \sqrt{17}}\right)$

$=\frac{\pi }{2}-{\cos }^{-1}\left(\sqrt{\frac{3}{17}}\right)$

$={\sin }^{-1}\left(\sqrt{\frac{3}{17}}\right) \left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right]$