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