Finding the parametric and vector forms of the line is perpendicular to lines $(4t,1+2t,3t)$ and $(-1+s,-7+2s,-12+3s)$

Question

Finding the parametric and vector forms of the line is perpendicular to the lines $$s_1(t) = (4t, 1+2t, 3t)$$ $$s_2(s) = (-1 + s, -7 + 2s, - 12 + 3s)$$

And passes through the point of the intersection of two lines $s_1$ and $s_2$

The vector form for $s_1$ is

$[x, y, z] = [0, 1, 0] + t[4, 2, 3]$

The parametric form is

$x = 4t, y = 1 + 2t, z = 3t$

The vector form for $s_2$ is (-1 + s, -7 + 2s, - 12 + 3s)$

$[x, y, z] = [-1, -7, -12] + s[1, 2, 3]$

The parametric form is

$x = -1 + s, y = -7 + 2s, z = -12 + 3s$

Not sure how to do this question

Answer 1

We find a vector $\vec{v}$ which is perpendicular to these lines, then $$\vec{v}=(4,2,3)\times(1,2,3)=(0,-9,6)$$ two lines meet in $(4,3,3)$, so the desired line is $$x=4~~~,~~~\dfrac{y-3}{-9}=\dfrac{z-3}{6}$$