Answer
$$\dfrac{2 A}{\sqrt 3}$$
Work Step by Step
We have: $curl F=3i+j-2k$
Need to re-write the equation as: $\int_{C} F \cdot dr=\int_{C} (z i-2 x j+3y k) \cdot (dx i+dy j+dz k)$
Let us suppose that $S$ be the part of the plane $x+y+z=1$ and this is the region enclosed by the loop $C$.
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr $
Now, $$\int_{C} F \cdot dr=\iint_{S} F \cdot dS= \iint curl F \cdot n dS \\=\dfrac{1}{\sqrt 3} \times \iint_{S}(3 i+j -2k) \cdot (i+j+k) dS \\=\dfrac{1}{\sqrt 3} \times \iint_{S}(3+1 -2) dS \\=\dfrac{2}{\sqrt 3} \iint_{S} dS \\=\dfrac{2 A}{\sqrt 3}$$
