Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$\begin{array}{rl}& L_1:\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ & \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda -1)\hat{\mathrm{i}}+2(\lambda +1)\hat{\mathrm{j}}+(\lambda +1)\hat{\mathrm{k}}\\ & {\mathrm{L}}_2:\overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ & \Rightarrow \overrightarrow{\mathrm{r}}=2\mu \hat{\mathrm{i}}+(1+7\mu )\hat{\mathrm{j}}+(1+3\mu )\hat{\mathrm{k}}\end{array}$


For point of intersection equating respective components
$\begin{array}{rl}& \Rightarrow \lambda -1=2\mu \\ & 2(\lambda +1)=1+7\mu \\ & \lambda +1=1+3\mu \end{array}$

We get
$\begin{array}{rl}& \Rightarrow \lambda =3\text{ and }\mu =1\\ & \Rightarrow \vec{a}+\vec{b}=3\hat{i}+9\hat{j}+4\hat{k}\end{array}$
${\mathrm{L}}_3:\overrightarrow{\mathrm{r}}=2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}}+\alpha (3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}})$

For $\alpha =2,\overrightarrow{\mathrm{r}}=8\hat{\mathrm{i}}+26\hat{\mathrm{j}}+12\hat{\mathrm{k}}$