From any point perpendiculars PM and PN are drawn to ZX and XY planes. If O is the origin, find the…

Since M is the foot of the perpendicular on the ZX plane, $M(2, 0, 2)$. Since N is the foot of the perpendicular on the XY plane, $N(2, 1, 0)$. Plane through origin will be $Ax + By + Cz = 0$. Since M and N lie on the plane, we get $\frac{A}{1} = \frac{B}{-2} = \frac{C}{-1}$. The equation of plane is $x - 2y - z = 0$.

d.r of OP $\langle 2, 1, 2\rangle$; d.r of normal to plane OMN $\langle \frac{1}{2}, -1, -\frac{1}{2}\rangle$ $\cos(90^\circ - \theta) = \frac{2\left(\frac{1}{2}\right) + 1(-1) + 2\left(-\frac{1}{2}\right)}{\sqrt{2^2+1+2^2}\sqrt{\left(\frac{1}{2}\right)^2+1+\left(\frac{1}{2}\right)^2}} = \frac{-2}{3\sqrt{6}}$ $\theta = \sin^{-1}\left(\frac{2}{3\sqrt{6}}\right)$ (as acute angle)

$\alpha = \sin^{-1}\left(\frac{2}{\sqrt{2^2+1+2^2}}\right) = \sin^{-1}\left(\frac{2}{3}\right)$ $\beta = \sin^{-1}\left(\frac{1}{\sqrt{2^2+1+2^2}}\right) = \sin^{-1}\left(\frac{1}{3}\right)$ $\gamma = \sin^{-1}\left(\frac{2}{\sqrt{2^2+1+2^2}}\right) = \sin^{-1}\left(\frac{2}{3}\right)$ R.H.S $= \csc^2\alpha + \csc^2\beta + \csc^2\gamma = \frac{9}{4} + 9 + \frac{9}{4} = \frac{27}{2} = \left(\frac{3\sqrt{6}}{2}\right)^2 = L.H.S.$