JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 14)

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+7 \hat{j}+3 \hat{k}$. Let $\mathrm{L}_1 : \overrightarrow{\mathrm{r}}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \vec{a}, \mathrm{\lambda} \in \mathbf{R}$ and $\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{j}+\hat{k})+\mu \vec{b}, \mu \in \mathrm{R}$ be two lines. If the line $\mathrm{L}_3$ passes through the point of intersection of $\mathrm{L}_1$ and $L_y$ and is parallel to $\vec{a}+\vec{b}$, then $L_3$ passes through the point :
$(-1, -1, 1)$
$(2, 8, 5)$
$(8, 26, 12)$
$(5, 17, 4)$

Explanation

$$\begin{aligned} & L_1: \overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda-1) \hat{\mathrm{i}}+2(\lambda+1) \hat{\mathrm{j}}+(\lambda+1) \hat{\mathrm{k}} \\ & L_2: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=2 \mu \hat{\mathrm{i}}+(1+7 \mu) \hat{\mathrm{j}}+(1+3 \mu) \hat{\mathrm{k}} \end{aligned}$$

For point of intersection equating respective components

$$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda-1=2 \mu \quad\text{....... (1)}\\ & 2(\lambda+1)=1+7 \mu \quad\text{...... (2)}\\ & \lambda+1=1+3 \mu \quad\text{..... (3)} \end{aligned}\\ &\text { We get } \end{aligned}$$

$$\begin{aligned} & \Rightarrow \lambda=3 \text { and } \mu=1 \\ & \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & L_3: \overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\alpha(3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ & \text { For } \alpha=2, \overrightarrow{\mathrm{r}}=8 \hat{\mathrm{i}}+26 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \end{aligned}$$

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