<s>[INST] Point of intersection when only direction ratios are given I am starting out with 3D Geometry. In one of the test booklets, I found a question for which I have no idea where and how to start from. 

If a line with direction ratio $2:2:1$ intersects the line $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$ and $\frac{x-1}{2}$ = $\frac{y+1}{4}$ = $\frac{z+1}{3}$ at A and B, find AB.

I know the algorithm for finding the points of intersection when a line is in the symmetrical form. I tried with an approach wherein I tried to find A by equating:
$\frac{x}{2}$ = $\frac{y}{2}$ = $\frac{z}{1}$ and $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$
However, on equating, I found out that these lines never even intersect.
How do I approach this sum? How to start off with this sum?
 [/INST] Let $a$ be the common value of $$\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{2}=a$$
We can re-write these equations under the following equivalent parametric form:
$$\tag{1}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7+3a\\5+2a\\3+2a\end{pmatrix}$$
In the same way, the generic point of the second straight line is :  $$\tag{2}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\ \ 1+2b\\-1+4b\\-1+3b\end{pmatrix}$$
Then you just have to express the proportionality of $\vec{AB}$ with the given vector, yielding the following system of 2 equations and 2 unknowns:
$$\dfrac{(1+2b)-(7+3a)}{2}=\dfrac{(-1+4b)-(5+2a)}{2}=\dfrac{(-1+3b)-(3+2a)}{1}$$
giving $a=-2/3$ and $b=1/3$. 
It remains to plug these values in (1) and (2) to get the coordinates of 

$$A=\begin{pmatrix}5\\11/3\\5/3\end{pmatrix} \ \ \ \text{and} \ \ \  B=\begin{pmatrix}5/3\\1/3\\0\end{pmatrix}.$$ 

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