QJEEM24Maff23: JEE Main 2024 Mathematics - Class 12 JEEM24Maff23 Solution: Let the image of the point in the line be the point . Then…

$\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$

$\begin{aligned}& \mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda) \\& \overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}\end{aligned}$

$\overrightarrow{\mathrm{PM}}$ is perpendicular to line $\mathrm{L}_1$

$\begin{aligned}& \overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\& \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 \\& 14 \lambda=14 \Rightarrow \lambda=1 \\& \therefore \mathrm{M}=(1,3,5) \\& \overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text { is midpoint of } \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}]\end{aligned}$

$\begin{aligned}& \overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\& \therefore(\alpha, \beta, \gamma)=(1,6,3)\end{aligned}$

Required line having direction cosine $(l, \mathrm{~m}, \mathrm{n})$

$\begin{aligned}& l^2+m^2+n^2=1 \\& \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1\end{aligned}$

$l^2=\frac{1}{4}$

$\therefore l=\frac{1}{2}$ [Line make acute angle with $\mathrm{x}$ -axis]

Equation of line passing through $(1,6,3)$ will be

$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$

Option (3) satisfying for $\mu=4$