Let \(\vec{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\vec{\mathrm{b}}=2\hat{\mathrm{i}}…

Given

\({L}_{1}:\vec{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ \Rightarrow \vec{\mathrm{r}}=(\lambda -1)\hat{\mathrm{i}}+2(\lambda +1)\hat{\mathrm{j}}+(\lambda +1)\hat{\mathrm{k}}\)

Next,

\({L}_{2}:\vec{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ \Rightarrow \vec{\mathrm{r}}=2\mu \hat{\mathrm{i}}+(1+7\mu )\hat{\mathrm{j}}+(1+3\mu )\hat{\mathrm{k}}\)

for point of intersection of \({L}_{1}\text{and}{L}_{2}\)

\(\lambda -1=2\mu \text{and}2(\lambda +1)=1+7\mu \\ \text{On solving, }\lambda =3\text{ and }\mu =1\\ \Rightarrow \vec{\mathrm{a}}+\vec{\mathrm{b}}=3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}}\\ \text{hence, }\\ {\mathrm{L}}_{3}:\vec{\mathrm{r}}=2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}}+\alpha (3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}})\\ \text{from options, we can see}{\mathrm{L}}_{3}\mathrm{passes}\\ \mathrm{through}(8,26,12)\)