Q: Does the line through $(3,4,7)$ in the direction $(2,3,-6)$ intersect the line passing through $(5,3,-2)$ and $(1,3,-9)$?
I said that the equation of the first line is $r_1=\begin{bmatrix}3\\4\\7\end{bmatrix}+s\begin{bmatrix}2\\3\\-6\end{bmatrix}=\begin{bmatrix}3+2s\\4+3s\\7-6s\end{bmatrix}$, and the equation of the second is $r_2=\begin{bmatrix}5-4t\\3\\-2-7t\end{bmatrix}$. In order for these to intersect, each row must agree with each other, so from row 2 we have $4+3s=3$ and so $s=-\frac{1}{3}$. Subbing into row 3 gives $t=-\frac{11}{7}$. Now substituting into the first row, we have $3+2s=\frac{7}{3}$ whereas $5-4t=\frac{79}{7}$ so the lines do not intersect. Is this correct?
Yes, you're right.
You can set these out using Left Hand Side and Right Hand side, solve the simultaneous equations using the two easiest rows and then check in the remaining row, e.g
$y:L.H.S = 32 , R.H.S. = 20$, therefore lines do not intersect, or
$y:L.H.S = 32 , R.H.S. = 32$, therefore lines intersect at...
and find the point by substituting $s$ or $t$ back into either line. If you hold two pencils up and look from the right angle, they'll seem to cross each other, but then if you move around, you might see a gap between them, that's the third row not matching, the lines are skew...or they can properly intersect in all 3 dimensions, or they can be parallel.