Let the equation of the plane be

Ax+By+Cz+D=0………..(i) (where D≠0)

As per the given criteria, the plane makes 3, -4, 2 intercepts on x, y, z axes respectively.

Hence the plane meets the x, y, z axes (3, 0, 0), (0, -4, 0) and (0, 0, 2) respectively.

Therefore by putting (0, 0, 2), we get

Similarly by putting (0, -4, 0) we get

And by putting (3, 0, 0) we get

Substituting the values of A, B, C in equation (i), we get

by putting B(0, -4, 0) we get

This is the Cartesian form of equation of the required plane

Now the vector equation of the plane 4x-3y+6z=12 can be written as

This is the required vector equation of the plane with intercepts 3, -4 and 2 on x, y and z axes respectively.