Given the distance between the point A(m,6) and the point B(3-m,2) is 10 units. Find the possible values of m.

That point will lie on the line through the origin that has a direction vector equal to the normal vector of the plane. Then the parametric equation for the line is
.. (x, y, z) = (0, 0, 0) +t*(3, 4, -1)

The point on this line that satisfies the plane equation will correspond to the t value that satisfies
.. 3(3t) +4(4t) -1(-t) = -4
.. t(9 +16 +1) = -4
.. t = -4/26 = -2/13

The point of interest is
.. (x, y, z) = (-6/13, -8/13, 2/13)