Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b}=2\hat{i}+7\hat{j}+3\hat{k}$. Let $L_1:\overrightarrow{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\overrightarrow{a}$, $\lambda\in\mathbf{R}$ and $L_2:\overrightarrow{r}=(\hat{j}+\hat{k})+\mu\overrightarrow{b}$, $\mu\in\mathbf{R}$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$, and is parallel to $\overrightarrow{a}+\overrightarrow{b}$, then $L_3$ passes through the point:
Step-by-Step Solution
Key Concept: Find the intersection of $L_1$ and $L_2$ by equating components and solving for $\lambda,\mu$, then write $L_3$ through that point with direction $\vec{a}+\vec{b}$ and verify which option lies on it.
$L_1$: $(\lambda-1,2\lambda+2,\lambda+1)$; $L_2$: $(2\mu,1+7\mu,1+3\mu)$.
$\lambda-1=2\mu$, $2\lambda+2=1+7\mu$, $\lambda+1=1+3\mu$.
From 3rd: $\lambda=3\mu$. Sub in 1st: $3\mu-1=2\mu \Rightarrow \mu=1$, $\lambda=3$.
Intersection: $(2,8,4)$.
$\vec{a}+\vec{b}=3\hat{i}+9\hat{j}+4\hat{k}$. $L_3$: $\vec{r}=2\hat{i}+8\hat{j}+4\hat{k}+\alpha(3\hat{i}+9\hat{j}+4\hat{k})$.
At $\alpha=2$: $(8,26,12)$ ✓.
Correct Answer: 3