If the lines are intersecting then
$(a_2^2-a_1).(b_1 \times b_2)=0$
$a_1=3\overrightarrow{i}+2\overrightarrow{j}-4\overrightarrow{k}$
$a_2=5\overrightarrow{i}-2\overrightarrow{j}$
$b_1=\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k}$
$b_2=3\overrightarrow{i}+2\overrightarrow{j}+6\overrightarrow{k}$
Therefore $a_2-a_1=2\overrightarrow{i}-4\overrightarrow{j}+4\overrightarrow{k}$
$\overrightarrow{b_1} \times \overrightarrow{b_2}=\begin{bmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{bmatrix}$
=$\overrightarrow{i}(12-4)-\overrightarrow{j}(6-6)+\overrightarrow{k}(2-6)$
=$8\overrightarrow{i}-0-4\overrightarrow{k}$
=$8\overrightarrow{i}-4\overrightarrow{k}$
Therefore $(a_2-a_1).(b_1 \times b_2)=(2\overrightarrow{i}-4\overrightarrow{j}+4\overrightarrow{k}).(8\overrightarrow{i}-4\overrightarrow{k})$
$=16-16=0$
Hence they are intersecting lines.
The position vectors of arbitrary points on the given lines are :
$L1 = 3i+2j-4k + \lambda(i+2j+2k)$
$= (3+\lambda)i+(2+2 \lambda)j+(-4+2 \lambda)k$
$L2=5i-2j+\mu(3i+2j+6k)$
$=(5+3\mu)i+(-2+2\mu)j+6\mu$
If they intersect they have a common point.
For some values of $\lambda$ and $\mu$ we must have $(3+\lambda)i+(2+2\lambda)j+(-4+2\lambda)k =(5+3\mu)i+(-2+2\mu)j+6\mu k$
$3+\lambda = 5+3\mu$
$2+2\lambda=-2+2\mu$
$ i.e., \lambda-3\mu=2$
$2\lambda-2\mu=-4$
on solving the two equations,
$\mu = -2\; and\; \lambda = -4$
Hence the point of intersection can be got by putting the value of lamda in the first line.
$ 3i+2j-4k+(-4)(i+2j+2k)$
$3i+2j-4k-4i-8j-8k$
$ -i-6j-12k$
Hence the coordinates of the points of intersection are (-1,-6,-12)