Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}}.$ Let ${\mathrm{L}}_1:\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{\lambda }\overrightarrow{\mathrm{a}},\mathrm{\lambda }\in \mathrm{R}$ and ${\mathrm{L}}_2:\overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{\mu }\overrightarrow{\mathrm{b}},\mathrm{\mu }\in \mathrm{R}$ be two lines. If the line L₃ passes through the point of intersection of L₁ and L₂, and is parallel to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$, then L₃ passes through the point:

$\begin{array}{l}{\mathrm{L}}_1:\overrightarrow{\mathrm{r}}=(-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\mathrm{k})+\mathrm{\lambda }(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\mathrm{k})\\ \Rightarrow \overrightarrow{\mathrm{r}}=(\mathrm{\lambda }-1)\widehat{\mathrm{i}}+2(\mathrm{\lambda }+1)\widehat{\mathrm{j}}+(\mathrm{\lambda }+1)\widehat{\mathrm{k}}\\ {\mathrm{L}}_2:\overrightarrow{\mathrm{r}}=(\widehat{\mathrm{j}}+\widehat{\mathrm{k}})+\mathrm{\mu }(2\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+3\widehat{\mathrm{k}})\\ \Rightarrow \overrightarrow{\mathrm{r}}=2\mathrm{\mu }\widehat{\mathrm{i}}+(1+7\mathrm{\mu })\widehat{\mathrm{j}}+(1+3\mathrm{\mu })\widehat{\mathrm{k}}\end{array}$
For point of intersection equating respective components
$\begin{array}{l}\Rightarrow \mathrm{\lambda }-1=2\mathrm{\mu }\\ 2(\mathrm{\lambda }+1)=1+7\mathrm{\mu }\\ \mathrm{\lambda }+1=1+3\mathrm{\mu }\end{array}$
We get
$\begin{array}{l}\Rightarrow \mathrm{\lambda }=3\text{ and }\mathrm{\mu }=1\\ \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3\widehat{\mathrm{i}}+9\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}\\ {\mathrm{L}}_3:\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{i}}+8\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}+\mathrm{\alpha }(3\widehat{\mathrm{i}}+9\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})\\ \text{For }\mathrm{\alpha }=2,\overrightarrow{\mathrm{r}}=8\widehat{\mathrm{i}}+26\widehat{\mathrm{j}}+12\widehat{\mathrm{k}}\end{array}$