The desired plane will have a direction vector that is perpendicular to that of the given line (4, -3, 1) and that of the given plane (2, 5, 4). Computing the cross product of these gives (-17, -14, 26).
Since the plane contains the point (-4, -3, -1), we can find the constant (d) in ax+by+cz=d using the dot-product of the point and the direction vector just found.
-17·(-4) -14·(-3) +26·(-1) = 68 +42 -26 = 84
Your plane can be written as
-17x -14y +26z = 84
Solving for z, you get
z = (17x +14y +84)/26
or, if you prefer, ...
z = (17/26)x +(7/13)y +42/13
