JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Three Dimensional Geometry

Points on the line can be written $D(3\lambda -2,1,4\lambda )$, and the foot of the perpendicular dropped from $A(1,-1,2)$ gives the triangle's height. The line's direction is $(3,0,4)$, and $\overrightarrow{AD}=(3\lambda -3,2,4\lambda -2)$ is perpendicular to it when $3(3\lambda -3)+0+4(4\lambda -2)=0$, i.e. $25\lambda =17$, so $\lambda =\frac{17}{25}$. Then $D=\left(\frac{1}{25},1,\frac{68}{25}\right)$ and the height is $AD=\sqrt{\frac{576}{625}+4+\frac{324}{625}}=\frac{2\sqrt{34}}{5}$. With base $BC=5$, the area is $\frac{1}{2}\times 5\times \frac{2\sqrt{34}}{5}=\sqrt{34}$.