$$\begin{aligned} & L: \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c} \\ & L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \text { (say) } \end{aligned}$$

Any point on $L_1$ be $A(2 \lambda+1,3 \lambda+1,4 \lambda+1)$

$$L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu \text { (say) }$$

Any point on $L_2$ be $B(\mu+3,2 \mu-4, \mu)$

$D$ of $L$ be : $<2 \lambda, 3 \lambda-2,4 \lambda>$ or $<\mu+2,2 \mu+3$, $\mu-1>$

Now $\frac{2 \lambda}{\mu+2}=\frac{3 \lambda-7}{2 \mu+3}=\frac{4 \lambda}{\mu-1}$

$$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda=\frac{-6}{5} \quad \mu=-5 \\ & \therefore< a, b, c >\equiv\langle-3,-7,-6 >\text { or }<3,7,6> \\ & \therefore \quad L: \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6} \end{aligned}\\ &(7,15,13) \text { lies on the line. } \end{aligned}$$