Find the equation of the plane passing through the line of intersection of the planes $\displaystyle \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=10$ and $\displaystyle \vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and passing through the point $\displaystyle (-2,3,1)$.

Equation of plane through the intersection of given two planes is \[\begin{aligned} & \vec{r} \cdot[(\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})]-10+4 \lambda=0 \\ & \vec{r} \cdot[(1+2 \lambda) \hat{i}+(1+3 \lambda) \hat{j}+(1-\lambda) \hat{k}]-10+4 \lambda=0 \end{aligned} \] Point (-$\displaystyle 2$, $\displaystyle 3$, $\displaystyle 1$) lies on it. \[\begin{aligned} & \therefore-2-4 \chi+3+9 \lambda+1-\lambda-10+4 \chi=0 \\ & 8 \lambda=8 \Rightarrow \lambda=1 \end{aligned} \] ∴ Equation of plane is \(\displaystyle \vec{r} .(3 \hat{i}+4 \hat{j})-6=0\)