Yes, those are parametric equations for a line with "t" as the parameter.
What pka is doing is constructing the line perpendicular to the given plane that goes through the given point. Where that line crosses the plane is the perpendicular projection of the point onto the plane.
-)))
Basically repeating what pka said, if the plane is given by Ax+ By+ Cz= D (equivalent to pka's "ax+ by+ cz= d" with A= a, B= b, C= c, D= -d) then < A, B, C> is a vector perpendicular to the plane. Any line perpendicular to the plane can be written, in parametric equations, as x= At+ x_0, y= Bt+ y_0, z= Ct+ z_0, where (x_0, y_0, z_0) is any point on the line. Since we want the point (m, n, j) to be on the line, x= At+ m, y= Bt+ n, z= Ct+ j.
To find the point where that line crosses the given plane, replace x, y, z in the equation of the plane with those parametric equations and solve for t.
For example- find the projection of (1, 3, 0) on the plane 2x+ y- z= 4.
A vector perpendicular to that plane is <2, 1, -1>. A line through (1, 3, 0) perpendicular to the plane is x= 2t+ 1, y= t+ 3, z= -t. Putting those into the equation of the plane, 2(2t+ 1)+ (t+ 3)- (-t)= 4t+ 2+ t+ 3+ t= 6t+ 5= 4. 6t= -1, t= -1/6. Putting that into the equation of the line, x= 2(-1/6)+ 1= 2/3, y= (-1/6)+ 3= 17/6, z= -(-1/6)= (1/6).
The perpendicular projection of the point (1, 3, 0) onto the plane 2x+ y- z= 4 is
(2/3, 17/6, 1/6).