If  $\overline{r}=\left(2-\lambda +\mu \right)\overline{i}+\left(1-\mu \right)\overline{j}+\left(2-3\lambda +2\mu \right)\overline{k}$ is the vector equation of a plane then the equivalent cartesian equation of the plane is

$\begin{array}{rl} & \text{Given}\overline{r}=(2-\lambda +\mu )\overline{i}+(1-\mu )\overline{j}+(2-3\lambda +2\mu )\overline{k}\\ & \Rightarrow \overline{r}=(2\overline{i}+\overline{j}+2\overline{k})+\lambda (-\overline{i}-3\overline{k})+\mu (\overline{i}-\overline{j}+2\overline{k})\end{array}$
$\text{which is a plane passing through}\overline{a}=2\overline{i}+\overline{j}+2\overline{k}\text{and}$
$\text{parallel to the vectors}\overline{b}=-\overline{i}-3\overline{k},\overline{c}=\overline{i}-\overline{j}+2\overline{k}$
$\text{It is perpendicular to the vector}\overline{n}\ne \overline{b}\times \overline{c}$
$=-3\overline{i}-\overline{j}+\overline{k}$
$\text{Hence vector equation is}(\overline{r}-\overline{a})\cdot \overline{n}=0$
$\begin{array}{rl} & \Rightarrow \overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}\\ & \Rightarrow -3x-y+z=-5\end{array}$
$\therefore \text{cartesian equation of plane is}3x+y-z=5$