JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Three Dimensional Geometry

The line $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ passes through $A(0,1,-1)$ with direction $(1,0,-1)$, and a general point on it is $C(\lambda ,1,-1-\lambda )$. Let $B=(\beta ,0,\beta )$. For $BC$ perpendicular to the line, $\overrightarrow{BC}\cdot (1,0,-1)=0$ gives $(\lambda -\beta )+(1+\lambda +\beta )=0$, i.e. $2\lambda +1=0$ and $\lambda =-\frac{1}{2}$, so $C=\left(-\frac{1}{2},1,-\frac{1}{2}\right)$. The foot distance is $|BC|=\sqrt{2{\left(\beta +\frac{1}{2}\right)}^2+1}$, and setting it equal to $\sqrt{\frac{3}{2}}$ gives $2{\left(\beta +\frac{1}{2}\right)}^2=\frac{1}{2}$, so ${\left(\beta +\frac{1}{2}\right)}^2=\frac{1}{4}$ and $\beta =0$ or $-1$. Since $\beta \ne 0$, $\beta =-1$.