The equation of the plane passing through $3 \hat{i}+2 \hat{j} +6 \hat{k}$ and parallel to the vectors $2 \hat{i}+ \hat{j}+ \hat{k}$ and $i-j+k$ is

Given that, the plane is parallel to the vectors $2 \hat{i}+ \hat{j}+\hat{k}$ and $\hat{i}- \hat{j}+\hat{k}$

So, $\overrightarrow{N}=(2 i+j+k)\times (i-j+k)$

$\overrightarrow{N}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{i}\\ 2& 1& 1\\ 1& -1& 1\end{array}\right|=2\hat{i}- \hat{j}-3\hat{k}$

$\overrightarrow{a}=3\hat{i}+2 \hat{j}+6\hat{k}$

Equation of the plane is, $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{N}=0$

$\Rightarrow \left(\right(x-3)\hat{i}+(y-2) \hat{j}+(z-6\left)\hat{k}\right).(2\hat{i}- \hat{j}-3\hat{k})=0$

$\Rightarrow 2(x-3)-(y-2)-3(z-6)=0$

$\Rightarrow 2x-y-3z=-14$