If the plane 2 x+y-5 z=0 is rotated about its line of intersection with the plane 3 x-y+4 z-7=0 by an angle of $\frac{\pi }{2}$, then the plane after the rotation passes through the point:

Equation of plane passing through the line of intersection of given planes is

$(2x+y-5z)+\lambda (3x-y+4z-7)=0$

$$\begin{gathered} (2+3\lambda )x+(1-\lambda )y+(-5+4\lambda )z-7\lambda =0 \\ this plane makes \frac{\pi }{2} with the plane \\ 2 x+y-5 z=0 \\ 2(2+3\lambda )+(1-\lambda )-5(-5+4\lambda )=0 \\ \Rightarrow 4+6\lambda +1-\lambda +25-20\lambda =0 \\ 30=15\lambda \\ \lambda =2 \end{gathered}$$

Required plane is  8 x-y+3 z-14=0

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