How to find vector representing altitude of a triangle?

Given the input vectors

   Vector3 AB =  Vector3(Bx, By, Bz) - Vector3(Ax, Ay, Az);
   Vector3 Caxis = Vector3(Cx, Cy, Cz) - Vector3(Ax, Ay, Az); // This is the vector from A to C

The direction of the result vector C we seek is the same direction as (AB cross Caxis) cross AB

   Vector3 Cdir = Vector3.Cross(Vector3.Cross(AB, Caxis), AB).Normalize();

Finding the magnitude is more involved. We know that starting at the point Cz and going in the direction Cdir must intersect along AB

   Caxis - alpha * Cdir == beta * AB

and we need to find unknowns alpha and beta. We can find a vector perpendicular to Caxis by projecting AB onto the plane to which Caxis is normal

   Vector3 Cperp = Vector3.ProjectOnPlane(Caxis, AB).Normalize();

so our equation with alpha and beta can be reduced to one unknown

  - alpha * (Cdir . Cperp)/(AB . Cperp) == beta

So solving for alpha (which is in fact the magnitude of our final vector C)

   Caxis == alpha * (Cdir - AB* (Cdir . Cperp)/(AB . Cperp) )

This is a vector equation, but we can solve for alpha by taking the dot product with Caxis on both sides. Writing code again this is

   float num = Caxis.magnitude*Caxis.magnitude;
   float denom1 = Vector3.Dot(Cdir, Caxis);
   float denom2 = Vector3.Dot(AB, Caxis) * Vector3.Dot(Cdir, Cperp) / Vector3.Dot(AB, Cperp);

   float Cmag = num/(denom1 - denom2); 

C is then, using the sign convention from your diagram,

   Vector3 C = Cmag*Cdir;

I'm leaving it without checks for cases where you get a divide by zero, such as when points B and C are identical.