Question: Use the big M method to solve

min z = 3X1

s.t. 2X1 + X2 >= 6
3X1 + 2X2 = 4
X1 , X2 >= 0


initial tableau

BV...... X1......X2.......e1......a1.......a2......RHS...
a1........ 2........1........-1......1........0........6.....
a2........3..........2........0.......0........1.... ....4.....
Z.........-3.........0........0......-M......-M.......0.....

(a1)M + (a2)M + Z

BV...... X1......X2.......e1......a1.......a2......RHS... Ratio
a1........ 2........1........-1......1........0........6.......3
a2........3..........2........0.......0........1.... ....4..... 4/3
Z.......5M-3......3M.....-M.......0........0.......10M... X1 enters, a2 leaves

BV...... X1......X2.......e1......a1.......a2..........RHS...
a1........0.......-1/3.....-1......1.......-2/3.......10/3...
X1........1.......2/3........0......0......1/3...........4/3.....
Z.........0....(-M/3)+2..-M......0...(-5M/3)+1..(10M/3)+4...

no more non-negative coefficients, can't get a1 out of basis, so solution is infeasable. This is what i did so far. If i have made any errors please tell me where i went wrong.