JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Three Dimensional Geometry

Take a point $(-2,-1,0)$ on the given line. The perpendicular from this point to the plane $x+y+z=3$ has direction ratios $(1,1,1)$ (normal to the plane). The line is $\frac{x+2}{1}=\frac{y+1}{1}=\frac{z}{1}=\lambda$, giving point $(\lambda -2,\lambda -1,\lambda )$. This lies on the plane when $(\lambda -2)+(\lambda -1)+\lambda =3$, so $3\lambda =6$, thus $\lambda =2$. The foot is $(0,1,2)$. Taking another point $(0,-2,3)$ on the original line, the perpendicular gives foot $(2/3,-4/3,11/3)$. The direction ratios of the line joining these feet are $(2/3,-7/3,5/3)$ or $(2,-7,5)$. The line through $(0,1,2)$ with these direction ratios is $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$.