JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Three Dimensional Geometry

For every point of $L$ to remain a fixed distance from each plane, $L$ must be parallel to both planes, so its direction is ${\vec{n}}_1\times {\vec{n}}_2=(1,2,-1)\times (2,-1,1)=(1,-3,-5)$, giving points $(t,-3t,-5t)$. Dropping a perpendicular from such a point onto $P_1$ means moving along ${\vec{n}}_1=(1,2,-1)$ by a parameter $s$ until $(t+s,-3t+2s,-5t-s)$ satisfies $x+2y-z+1=0$; substituting gives $6s+1=0$, so $s=-\frac{1}{6}$. The foot is therefore $(t-\frac{1}{6},-3t-\frac{1}{3},-5t+\frac{1}{6})$, which traces the locus $M$ as $t$ varies. Taking $t=0$ gives $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$ and $t=\frac{1}{6}$ gives $(0,-\frac{5}{6},-\frac{2}{3})$, both of which appear among the listed points, whereas no value of $t$ reproduces $(-\frac{5}{6},0,\frac{1}{6})$ or $(-\frac{1}{3},0,\frac{2}{3})$.