• # question_answer If $Q\left( 0,1,3 \right)$is the image of the point P in the plane $3xy+4z=2$and R is the point (3, -1, -2), then the area (in sq. units) of $\Delta PQR$is: [JEE Main 10-4-2019 Morning] A) $\frac{\sqrt{65}}{2}$                       B) $\frac{\sqrt{91}}{4}$C) $2\sqrt{13}$              D) $\frac{\sqrt{91}}{2}$

R lies on the plane.             $DQ=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}=\sqrt{\frac{13}{2}}$$\Rightarrow PQ=\sqrt{26}$ Now,$RQ=\sqrt{9+1}=\sqrt{10}$ $\Rightarrow RD=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}$ Hence,$ar(\Delta PQR)=\frac{1}{2}\times \sqrt{26}\times \sqrt{\frac{7}{2}}=\sqrt{\frac{91}{2}}.$