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

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

$$\overrightarrow{\mathrm{PM}}$$ 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 $$\mu=4$$