Find the distance of the point ($\displaystyle 1$, -$\displaystyle 2$, $\displaystyle 9$) from the point of intersection of the line $\displaystyle \vec{\mathrm{r}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ and the plane $\displaystyle \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=10$. Case Study Problem :

Given line is \(\displaystyle \vec{r}=(4 \hat{\imath}+2 \hat{\jmath}+7 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k})\) Any point on the line is \[(4+3 \lambda, 2+4 \lambda, 7+2 \lambda) \] This point lies on the plane \(\displaystyle \vec{r} .(\hat{\imath}-\hat{\jmath}+\widehat{k})=10\) \[\begin{aligned} & \Rightarrow 4+3 \lambda-2-4 \lambda+7+2 \lambda=10 \\ & \Rightarrow \lambda+9=10 \Rightarrow \lambda=1 \end{aligned} \] ∴ Point of intersection is \(\displaystyle (7,6,9)\) Distance of ($\displaystyle 7$, $\displaystyle 6$, $\displaystyle 9$) from ($\displaystyle 1$, -$\displaystyle 2$, $\displaystyle 9$) is \[\begin{aligned} & d=\sqrt{(7-1)^{2}+(6+2)^{2}+0} \\ & =\sqrt{36+64}=10 \end{aligned} \]