Solution

From $-3x_1 + x_2 \ge 3$

$x_2 \ge 3 + 3x_1$

But from $x_1 + x_2 \le 1$

$x_2 \le 1 - x_1$

So we need:

$3 + 3x_1 \le 1 - x_1$

$4x_1 \le -2$

$x_1 \le -\frac{1}{2}$

This contradicts $x_1 \ge 0$

Hence problem is infeasible.