The distance of the point with position vector $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} . (-2 \hat{i}+\hat{j}+3 \hat{k})=8$ is :

$\sqrt{14}$

Point → $(2 \hat{i}+\hat{j}-\hat{k}) ≡ P(2, 1, -1)$

Plane → $\vec{r}(-2 \hat{i} + \hat{j} + 3\hat{k}) = 8$

⇒ $(x \hat{i}+y \hat{i}+z \hat{k})(-2 \hat{i}+\hat{j}+3 \hat{k})-8=0$

$\Rightarrow -2 x+y+3 z-8=0$ → equation of plane

as for point (x₀, y₀, z₀) and plane

Ax + By + Cz + D = 0

$d=\frac{\left|A x_0+B y_0+C z_0+D\right|}{\sqrt{A^2+B^2+C^2}}$

so distance

$=\frac{|-2(2)+(1)+3(-1)-8|}{\sqrt{(-2)^2+1^2+3^2}}$

$=\frac{14}{\sqrt{14}}$

$=\sqrt{14}$