Vtyjkyuk

The vector $l_1$ is given as
$$
l_1 : mathbfr =
left(
beginmatrix
-3\-4\6
endmatrix
right); + lambda
left(
beginmatrix
-3\-4\6
endmatrix
right)
$$
and $l_2$ as
$$
l_2: fracx-4-3=fracy+74=-(z+3)
$$

So far I've gotten an equation for each $(x,y,z)$ of $l_1$ and $l_2$ and then multiplied them to get a scalar product however I'm having issues finding the values of $lambda$ and $nu$ (Cartesian multiplier of $l_2$) through the simultaneous equations of finding the scalar product (when it's equal to $0$).

My approach, at least so far has given,
$$
vecOB = left(beginmatrix
4\
-7\
-3
endmatrixright)+nu
left(beginmatrix
-3\
4\
-1
endmatrixright)
$$
$$
vecAB = vecOB-vecAB
$$
Therefore
$$
vecAB = left(beginmatrix
1\
3\
9
endmatrixright)+
left(beginmatrix
-3nu - 3lambda\
4nu - 2lambda\
-nu + 2lambda
endmatrixright)
$$
$$
x = 1 -3nu - 3 lambda,; y = -3 + 4nu - 2lambda, ; z=-9+nu + 2lambda
$$
Then solving for the two dot products,
$$
vecAB ; cdot ; l_1, ; vecAB ; cdot ; l_2
$$
I get
$$
26nu^2-17lambda^2 = 0
$$
Which I have no idea how to solve. The response was too long for the comment section.

A (slightly) different approach:

$l_2$ is defined as the intersection of the planes $x-3z=7$ and $y+4z=-19$, with normal vectors $vec n=(1,0,-3)$ and $vec n'=(0,1,4)$ respectively. Therefore, a directing vector of $l_2$ is the cross-product $vec u_2=vec ntimesvec n'$.

Let $vec u_1=(-3,-4,6)$ the given directing vector of $l_1$. A directing vector of the common perpendicular of $l_1$ and $l_2$ is the cross-product
$$vec v=vec u_1timesvec u_2. $$
Thus, to find points $A$ on $l_1$ and $B$ on $l_2$ so that the line $(AB)$ is perpendicular to $l_1$ and $l_2$, we have to find a point $A$ (given by the parametric equation) and $t$ so the point
$$B=A+tvec v= A+t(vec u_1timesvec u_2)$$
satisfies the equations of line $l_2$.