The Cartesian equation of the plane ${\overrightarrow{r}=(1+\lambda -\mu )\widehat{i}+(2-\lambda )\widehat{j}+(3-2\lambda +2\mu )\widehat{k}\text{ is }}_{ }$

Given plane is
$\begin{array}{r}\overrightarrow{r}=(1+\lambda -\mu )\widehat{i}+(2-\lambda )\widehat{j}+(3-2\lambda +2\mu )\widehat{k}\\ =(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda (\widehat{i}-\widehat{j}-2\widehat{k})+\mu (-\widehat{i}+2\widehat{k})\end{array}$

which is a plane passing through  $\overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}$

and parallel to the vectors $\overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k}$  and $\overrightarrow{c}=-\widehat{i}+2\widehat{k}$

Therefore, it is perpendicular to the vector
$\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}=-2\widehat{\mathrm{i}}-\widehat{\mathrm{k}}$

Hence, equation of plane is - 2(x - 1) + (0)(y - 2)(z-3) = 0  or 2x+z = 5