VITEEE 2021 Paper

Let us suppose that the given line and plane intersect at the point $P(x,y,z)$ . $\therefore$ The position vector of $P$ is $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$Thus, the given equations of the line and the plane can be rewritten as $x\hat{i}+y\hat{j}+z\hat{k}=(2+3\lambda)\hat{i}-(1+4\lambda)$ hayj $+(2+2\lambda)\hat{k}$ and $(x\hat{i}+y\hat{j}+z\hat{k})\cdot(\hat{i}-\hat{j}+\hat{k})=5$ , respectively.On simplifying $x\hat{i}+y\hat{j}+z\hat{k}=(2+3\lambda)\hat{i}-(1+4\lambda)\hat{j}+(2+2\lambda)\hat{k}$ and $(x\hat{i}+y\hat{j}+z\hat{k})\cdot(\hat{i}-\hat{j}+\hat{k})=5$ , we get:Given equation of line is $r=2\hat{i}-\hat{j}+2\hat{k}+\lambda(3\hat{i}+4\hat{j}+2\hat{k})$ $(x\hat{i}+y\hat{j}+z\hat{k})=(2+3\lambda)\hat{i}+(-1+4\lambda)\hat{j}+(2+2\lambda)\hat{k}$ Any point on the line is $(2+3\lambda,-1+4\lambda,2+2\lambda)$ Since it also lie on the plane $r \cdot (\hat{i}-\hat{j}+\hat{k})$⇒ $2+3\lambda+1-4\lambda+2+2\lambda=5 \Rightarrow \lambda=0$ Therefore, coordinate of the point of intersection of line and plane is $(2,-1,2)$ . ∴ Distance $d=\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}=13$