Find the vector equation of the line passing through $(1,2,3)$ and parallel to the planes $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6$.
Let the required line be parallel to vector 
given by,
$\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$
The position vector of the point $(1,2,3)$ is $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$
The equation of line passing through $(1,2,3)$ and parallel to $\vec{b}$ is given by,
$\vec{r}=\vec{a}+\lambda \vec{b}$
$\Rightarrow \vec{r}(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$ $\ldots(1)$
The equations of the given planes are
$\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5$ $\ldots(2)$
$\vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6$ $\ldots(3)$
The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.
$\Rightarrow(\hat{i}-\hat{j}+2 \hat{k}) \cdot \lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)=0$
$\Rightarrow \lambda\left(b_{1}-b_{2}+2 b_{3}\right)=0$
$\Rightarrow b_{1}-b_{2}+2 b_{3}=0$ $\ldots(4)$
Similarly, $(3 \hat{i}+\hat{j}+\hat{k}) \cdot \lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)=0$
$\Rightarrow \lambda\left(3 b_{1}+b_{2}+b_{3}\right)=0$
$\Rightarrow 3 b_{1}+b_{2}+b_{3}=0$ $\ldots(5)$
From equations (4) and (5), we obtain
$\frac{b_{1}}{(-1) \times 1-1 \times 2}=\frac{b_{2}}{2 \times 3-1 \times 1}=\frac{b_{3}}{1 \times 1-3(-1)}$
$\Rightarrow \frac{b_{1}}{-3}=\frac{b_{2}}{5}=\frac{b_{3}}{4}$
Therefore, the direction ratios of $\vec{b}$ are $-3,5$, and $4 .$
$\therefore \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}=-3 \hat{i}+5 \hat{j}+4 \hat{k}$
Substituting the value of $\vec{b}$ in equation (1), we obtain
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})$
This is the equation of the required line.