$$P$$ is a point on the $$y$$–$$z$$ plane, making equal angles with the $$y$$-axis and $$z$$-axis and at a distance $$2$$ from the origin. $$M$$ is the foot of the perpendicular from $$P$$ to the plane $$3x + y - \sqrt 2 z = 2\sqrt 2 .$$     The coordinates of $$M$$ are :

Solution :
$$\eqalign{ & P = \left( {0,\,2\cos \,{{45}^ \circ },\,2\sin \,{{45}^ \circ }} \right) = \left( {0,\,\frac{1}{{\sqrt 2 }},\,\frac{1}{{\sqrt 2 }}} \right). \cr & {\text{If }}M = \left( {\alpha ,\,\beta ,\,\gamma } \right){\text{ then}} \cr & 3\alpha + \beta - \sqrt 2 \gamma \, = 2\sqrt 2 {\text{ and }}\frac{{\alpha - 0}}{3} = \frac{{\beta - \frac{1}{{\sqrt 2 }}}}{1} = \frac{{\gamma - \frac{1}{{\sqrt 2 }}}}{{ - \sqrt 2 }} \cr & \therefore \frac{\alpha }{3} = \frac{{\beta - \frac{1}{{\sqrt 2 }}}}{1} = \frac{{\sqrt 2 \gamma - 1}}{{ - \sqrt 2 }} \cr & = \frac{{3\alpha + \beta - \frac{1}{{\sqrt 2 }} - \left( {\sqrt 2 \gamma - 1} \right)}}{{9 + 1 - \left( { - 2} \right)}} \cr & = \frac{{2\sqrt 2 - \frac{1}{{\sqrt 2 }} + 1}}{{12}} \cr & = \frac{{3 + \sqrt 2 }}{{12\sqrt 2 }} \cr & \therefore \,\,\alpha = \frac{{3 + \sqrt 2 }}{{4\sqrt 2 }}, \cr & \beta = \frac{1}{{\sqrt 2 }} + \frac{{3 + \sqrt 2 }}{{12\sqrt 2 }} = \frac{{15 + \sqrt 2 }}{{12\sqrt 2 }}, \cr & \gamma = \frac{1}{{\sqrt 2 }}\left\{ {1 - \frac{{6 + 2\sqrt 2 }}{{12\sqrt 2 }}} \right\} = \frac{{10\sqrt 2 - 6}}{{24}} \cr} $$