Your parameterization is not correct.
A parameterization of a line has the form $r(t)= P + t D$ where $P$ is a vector "touching" the line and $D$ is a direction vector for the line.
Given two points on the line, $P$ and $Q$, the equation $r(t) = P+ t Q$ is not the correct parameterization. This is where you are in error. So, instead of using $Q$ as you did, use the displacement vector from $P$ to $Q$. In your case, this would be
$$
Q-P = (-2,1,1) - (3,-1,1) =\bigl(-2-3 , 1-(-1), 1-1 \bigr)=(-5,2,0).
$$
The parameterization is then
$$
r(t)=(3,-1,1)+t(-5,2,0).
$$
(And don't forget to give a range of values for $t$.)
Your procedure for finding the symmetric equations looks ok.
I think a more fundamental way to approach the problem is by discussing geodesic curves on the surface you call home. Remember that the geodesic equation, while equivalent to the Euler-Lagrange equation, can be derived simply by considering differentials, not extremes of integrals. The geodesic equation emerges exactly by finding the acceleration, and hence force by Newton's laws, in generalized coordinates.
See the Schaum's guide Lagrangian Dynamics by Dare A. Wells Ch. 3, or Vector and Tensor Analysis by Borisenko and Tarapov problem 10 on P. 181
So, by setting the force equal to zero, one finds that the path is the solution to the geodesic equation. So, if we define a straight line to be the one that a particle takes when no forces are on it, or better yet that an object with no forces on it takes the quickest, and hence shortest route between two points, then walla, the shortest distance between two points is the geodesic; in Euclidean space, a straight line as we know it.
In fact, on P. 51 Borisenko and Tarapov show that if the force is everywhere tangent to the curve of travel, then the particle will travel in a straight line as well. Again, even if there is a force on it, as long as the force does not have a component perpendicular to the path, a particle will travel in a straight line between two points.
Also, as far as intuition goes, this is also the path of least work.
So, if you agree with the definition of a derivative in a given metric, then you can find the geodesic curves between points. If you define derivatives differently, and hence coordinate transformations differently, then it's a whole other story.
Best Answer
No, there is no way. $$ \frac{\lVert AB \times AP \rVert}{\lVert AB \rVert} = \frac{\lVert AB \rVert \lVert AP \rVert \sin \alpha}{\lVert AB \rVert} = \lVert AP \rVert \sin \alpha $$ This formula only depends on the distance between $A$ and $P$ and that angle between the first line and $AP$ which is the shortest connection between $A$ and $P$, so it leaves out the direction of the second line.
We could rotate the second line around $P$ without changing the value of that formula, while that rotation might change the minimal distance between both lines.
For an example minimizing the (squared) distance over two parameters see here.