The vector equation of the plane through the point $\hat{i}+2\hat{j}-\hat{k}$  and $\perp$  to the line of intersection of the plane $\overrightarrow{r}.\left(3\hat{i}-\hat{j}+\hat{k}\right)=1$  and $\overrightarrow{r}.(\hat{i}+4\hat{j}-2\hat{k})=2$  is

detailed solution

Correct option is B

The line of intersection of the planes r→.(3i^−j^+k^)=1  and r→.(i^+4j^−2k^)=2   is common to both the planes. Therefore, it is ⊥  to normals to the two planes i.e. n→1=3i^−j^+k^  and n→2=i^+4j^−2k^ . Hence it is parallel to the vector n→1×n→2=−2i^+7j^+13k^ . Thus, we have to find the equation of the plane passing through a→=i^+2j^−k^  and normal to the vector n→=n1→×n2→ . The equation of the required plane is(r→−a→).n→  or r→.n→=a→.n→ or r→.(−2j^+7j^+13k^)=(i^+2j^−k^).(−2j^+7j^+13k^)  or r→.(2i^−7j^−13k^)=1