JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Three Dimensional Geometry

The direction ratios of the line (intersection of two planes) are given by ${\vec{n}}_1=\hat{i}+2\hat{j}+3\hat{k}$ and ${\vec{n}}_2=2\hat{i}+\hat{j}-\hat{k}$. Their cross product gives the direction of the line: ${\vec{n}}_1\times {\vec{n}}_2=-5\hat{i}+7\hat{j}-3\hat{k}$. The normal to the second plane is ${\vec{n}}_3=(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}-2\hat{j}+3\hat{k})=5\hat{i}-2\hat{j}-3\hat{k}$. The required plane contains the line and is perpendicular to the second plane, so its normal is $(-5\hat{i}+7\hat{j}-3\hat{k})\times (5\hat{i}-2\hat{j}-3\hat{k})=-27\hat{i}-30\hat{j}-25\hat{k}$. A point on the line can be found by solving the system; one such point is $(0,-\frac{11}{5},\frac{14}{5})$. The plane equation is $27x+30y+25z=4$, so $a=27,b=30,c=25$, giving $a-b+c=22$.