The given line is $\vec{r}=2\hat{i}-\hat{j}+2\hat{k}$ $+\lambda(3\hat{i}+4\hat{j}+2\hat{k})$ ...(1) and the given plane is $\vec{r}\cdot(\hat{i}-\hat{j}+\hat{k})=5$...(2) Solving (1) and (2), we get $(2+3\lambda)(1)+(4\lambda-1)(-1)+(2\lambda+2)(1)=5$$\Rightarrow 2+3\lambda-4\lambda+1+2\lambda+2=5$ $\Rightarrow \lambda=0$∴ The point of intersection of (1) and (2) is $(2\hat{i}-\hat{j}+2\hat{k})$ i.e., (2 1− 2)The other point is (–1, –5, –10).∴ Required distance $=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}=13$