Any point on the first line in symmetrical form is $(3\mathrm{r}-4, 5\mathrm{r}-6, -2\mathrm{r}+1)$. If the lines are coplanar, this point must lie on both the planes which determine the second line. $\Rightarrow 3(3\mathrm{r}-4)-2(5\mathrm{r}-6)-2\mathrm{r}+1+5=0 \dots .\text{ (i) }$

and $2(3\mathrm{r}-4)+3(5\mathrm{r}-6)+4(-2\mathrm{r}+1)-\mathrm{k}=0 \dots .\left(\mathrm{ii}\right)$

From Eq. (i) we get, r = 2.

Now substituting r = 2 in Eq. (ii), then k = 4