Line $L_1$ is passing through $5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ and parallel to the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$.
$$
\therefore \quad L_1 \equiv \mathbf{r}=5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})
$$
line $L_2$ is passing through $4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$ and parallel to the vector $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
$$
\therefore \quad L_2 \equiv \mathbf{r}=4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})
$$
$L_1$ and $L_2$ are intersecting.
$$
\begin{aligned}
& \therefore \quad 5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}} \\
& +8 \hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \\
& \Rightarrow \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=(3 \mu-2 \lambda) \hat{\mathbf{i}}+(4 \mu-3 \lambda) \hat{\mathbf{j}}+(5 \mu-4 \lambda) \hat{\mathbf{k}} \\
& \Rightarrow \quad 3 \mu-2 \lambda=1 \\
& 4 \mu-3 \lambda=2 \\
&
\end{aligned}
$$
On solving Eqs. (i) and (ii), we get

$$
\mu=-1, \lambda=-2
$$
Put $\quad \lambda=-2 \operatorname{in} L_1$
We get $\quad r=\hat{i}+2 \hat{\mathbf{j}}+3 \hat{k}$
$\therefore$ Point of intersection of $L_1$ and $L_2$ is $\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathbf{k}}$