The distance of the point $(1,2,3)$ from the plane $\overrightarrow{r}.(2\widehat{i}+\widehat{j}+2\widehat{k})+5=0$ measured parallel of the line $\overrightarrow{r}=(-3\widehat{i}+2\widehat{j})+\lambda (\widehat{i}+\widehat{j}+\widehat{k})$ is

Given plane is  $r.(2i+j+2k)+5=0 \Rightarrow 2x+y+2z+5=0$
Given line passes through $(1,2,3)$ and parallel to $\overline{r}=(-3i+2j)+\lambda (i+j+k)$  is  $\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=t$

Any point on the line $Q(t+1,t+2,t+3)$ lies on plane

$$\begin{gathered} 2(t+1)+t+2+2(t+3)+5=0\Rightarrow t=-3 \\ \therefore Q(-2,-1,0)P(1,2,3) \\ \therefore PQ=\sqrt{9+9+9}=3\sqrt{3} \end{gathered}$$