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