Find the vector equation of the plane passing through the intersection of the planes

The equations of the planes are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$

$\Rightarrow \vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7=0$

$\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$[\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7]+\lambda[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]=0$, where $\lambda \in \mathbf{R}$

$\vec{r} \cdot[(2 \hat{i}+2 \hat{j}-3 \hat{k})+\lambda(2 \hat{i}+5 \hat{j}+3 \hat{k})]=9 \lambda+7$

$\vec{r} \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(3 \lambda-3) \hat{k}]=9 \lambda+7$                  ..(3)

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\vec{r}=2 \hat{i}+1 \hat{j}+3 \hat{k}$

Substituting in equation (3), we obtain

$(2 \hat{i}+\hat{j}+3 \hat{k}) \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(3 \lambda-3) \hat{k}]=9 \lambda+7$

$\Rightarrow 2(2+2 \lambda)+1(2+5 \lambda)+3(3 \lambda-3)=9 \lambda+7$

$\Rightarrow 4+4 \lambda+2+5 \lambda+9 \lambda-9=9 \lambda+7$

$\Rightarrow 18 \lambda-3=9 \lambda+7$

$\Rightarrow 9 \lambda=10$

$\Rightarrow \lambda=\frac{10}{9}$

Substituting $\lambda=\frac{10}{9}$ in equation (3), we obtain

$\vec{r} \cdot\left(\frac{38}{9} \hat{i}+\frac{68}{9} \hat{j}+\frac{3}{9} \hat{k}\right)=17$

$\Rightarrow \vec{r} \cdot(38 \hat{i}+68 \hat{j}+3 \hat{k})=153$

This is the vector equation of the required plane.