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

Solution:

The given planes are

$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$

$\Rightarrow \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1=0$

$\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$

The equation of any plane passing through the line of intersection of these planes is

$[\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1]+\lambda[\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\widehat{k})+4]=0$

$\vec{r} \cdot[(2 \lambda+1) \hat{i}+(3 \lambda+1) \hat{j}+(1-\lambda) \hat{k}]+(4 \lambda-1)=0 \quad \ldots(1)$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

$\therefore 1 .(2 \lambda+1)+0(3 \lambda+1)+0(1-\lambda)=0$

$\Rightarrow 2 \lambda+1=0$

$\Rightarrow \lambda=-\frac{1}{2}$

Substituting $\lambda=-\frac{1}{2}$ in equation (1), we obtain

$\Rightarrow \vec{r} \cdot\left[-\frac{1}{2} \hat{j}+\frac{3}{2} \hat{k}\right]+(-3)=0$

$\Rightarrow \vec{r}(\hat{j}-3 \hat{k})+6=0$