Answer
$\dfrac{-5}{6}$
Work Step by Step
Stoke's Theorem states that
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
We are given that vector field as
$F=yi+xzj+x^2k$
and the surface as
$x+y+z=1$
Now, $\oint F \cdot dr=\int_0^{1} \int_0^{1-y} (-xi-2xj+z-k) (\dfrac{(i+j+k)}{\sqrt 3}) \sqrt 3 dA = \int_0^{1}\int_0^{1-y} -4x-y \ dx dy \\= \int_0^1 [-2x^2+y^2x]_0^{1-y} \ dy \\=\int_0^1 -2(1+y^2-2y_ -y -y^2 \ dy \\=\int_0^1 -2+3y-y^2 \ dy \\=\dfrac{-1}{3} +\dfrac{3}{2} -2 \\=\dfrac{-5}{6}$
