Let the helix be given by $(\cos t, \sin t, ht)$ (after scaling). If $P$ is your point $(a,b,c)$, and $Q = (\cos t, \sin t, ht)$ is the nearest point on the helix, then $PQ$ is perpendicular to the tangent at $Q$, which is just $(-\sin t, \cos t, h)$:
$-(\cos t - a)\sin t + (\sin t - b)\cos t + (ht - c)h = 0 $
This simplifies to $A \sin(t+B) + Ct + D = 0$ for some constants $A,B,C,D,$ as Moron said. But then you have to solve this numerically. There will be more than one solution in general, but (as Jonas Kibelbek pointed out in the comments) you only need to check the solutions with $z$-coordinate in the interval $[c-\pi h, c+\pi h)$.
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
The area of the parallelogram spanned by points $A,B$ (on the line), and $C$ is $$|(B-A)\times (C-A)|=|(x_B-x_A)(y_C-y_A)-(y_B-y_A)(x_C-x_A)|.$$ If we divide this by the length $\sqrt{(B-A)^2}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ of its base, we obtain ist height.
Final formula: $$\frac{|(x_B-x_A)(y_C-y_A)-(y_B-y_A)(x_C-x_A)|}{\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}$$
So in your concrete example, the distance is $$ \frac{|(10-(-7))(6-9)-(9-9)(4-(-7))|}{\sqrt{(10-(-7))^2+(9-9)^2}}=3.$$