Let
L = x+y+1=0 and it has slope, m = -1
And let
L1 = 2x-3y+4=0 and it has slope, m1= 2/3
Let L2 line will be passing through intersection of Line L1, therefore
L2 = (2x - 3y +4) + k(x+y+1) = 0
= (2 + k)x + (k - 3)y + ( 4 + k) = 0
And let the slope for this line be, m2 = (2+k)/(3-k)
Now,
If theta is the angle between bisector and any of the other twolines then,
tan(theta) = (m2 - m)/(1+ mm2) = (m - m1)/(1+ mm1)

({(2 + k)/(3 - k)} + 1)/ (1 - {(2+k)/(3-k)})= (-1-2/3)/(1-2/3)

Now after solving your will get k=1
More putting this k in L2 to get the required equation of line which is 3x - 2y + 5 = 0
So the equation of other line is 3x - 2y +5 = 0