Take any point N(t- 1, t+3, 2t) on a known straight line,

Because the straight line passes through m (- 1, 0,4), the direction vector of the straight line MN is MN=(t, t+3,2t-4).

Because the straight line MN// plane, and the normal vector of the plane is n=(3, -4, 1),

So n*MN=0, that is, 3t-4(t+3)+2t-4=0, and the solution is t= 16.

Therefore, the linear direction vector is MN=( 16,19,28).

Then the equation is (x+1)16 = y/19 = (z-4)/28.