The vector equation of a plane through the inter- section of the planes $\vec{r}. (\hat{i} + 3\hat{j} - \hat{k})= 5 $ and $\vec{r}. (2\hat{i} - \hat{j} + \hat{k})= 3 $ and passing through the point (2, 1, -2), is

$\vec{r}. (3\hat{i} + 2\hat{j})= 8 $

The equation of a plane through the intersection of 

$\vec{r}. (\hat{i} + 3\hat{j} - \hat{k})= 5 $ and $\vec{r}. (2\hat{i} - \hat{j} + \hat{k})= 3 $ is given by

$[\vec{r}. (\hat{i} + 3\hat{j} - \hat{k})- 5] +λ[\vec{r}. (2\hat{i} - \hat{j} + \hat{k})- 3]=0 $

$⇒\vec{r} [(1+2λ)\hat{i} + (3- λ)\hat{j} + (-1+λ) \hat{k}]-5 - 3λ = 0 $ .....(i)

If (i) passes through (2, 1, -2), then the vector $2\hat{i} + \hat{j}-2\hat{k}$ should satisfy it.

$∴ (2\hat{i} + \hat{j} -2 \hat{k}).[(1+2λ)\hat{i}+(3-λ)\hat{j} + (-1 +λ)\hat{k}]- (5 + 3λ)=0 $

$ ⇒ 2(1 + 2λ)+ 1(3-λ) -2(-1 + λ) - (5 + 3 λ) = 0 $

$ ⇒ -2λ + 2 = 0 ⇒ λ = 1 $

Putting λ =1 in (i), we get the required equation of the plane as $\vec{r}. (3\hat{i} + 2\hat{j} +0 \hat{k})=8$.