Answer
$\oint_L \vec{v}\cdot \vec{dl} = \oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da}= \frac{8}{3}$
Work Step by Step
$\vec{v} = 6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k}$
$\vec{dl} = dx\hat{i} + dy\hat{j} + dz \hat{k}$
Break the line integral into 3 parts where
$L_1$ denotes the line integral where y changes from $0$ to $1$, $x=0, z=0$
$\int_{L_1}\vec{v}\cdot \vec{dl} =\int_{L_1} (6\hat{i} + 3y\hat{k})\cdot (dy\hat{j}) = 0$
$L_2$ denotes the line integral of the line $z = -2y+2, x=0$
$dz = -2dy$
$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k})\cdot (dy\hat{j} + dz\hat{k})$
$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (yz^2dy + (3y+z)dz$
$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (y(-2y+2)^2dy + (3y+(-2y+2))(-2dy)$
$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_1^0 (4y^3 - 8y^2+2y-4)dy$
$\int_{L_2}\vec{v}\cdot \vec{dl} =\big[y^4 - \frac{8y^3}{3}+y^2-4y \big]_1^0= \frac{8}{3} + 2$
$L_3$ denotes the line integral where $z$ changes from $2$ to $0$, $x=0, y=0$
$\int_{L_3}\vec{v}\cdot \vec{dl} =\int_{L_3} (6\hat{i} + z\hat{k})\cdot (dz\hat{k}) $
$\int_{L_3}\vec{v}\cdot \vec{dl} =\int_2^0 z dz = -2 $
$\oint_L \vec{v}\cdot \vec{dl} = \int_{L_1} \vec{v}\cdot \vec{dl} + \int_{L_2} \vec{v}\cdot \vec{dl} + \int_{L_3} \vec{v}\cdot \vec{dl} = \frac{8}{3}$
$\vec{\triangledown} \times \vec{v} =
\begin{vmatrix}
\hat{i} && \hat{j} && \hat{k}\\
{\partial \over \partial x} && {\partial \over \partial y} && {\partial \over \partial z}\\
6 && yz^2 && (3y+z)
\end{vmatrix} $
$\vec{\triangledown} \times \vec{v} = \hat{i}(3-2yz) -0\cdot\hat{j} + 0\cdot\hat{k} = \hat{i}(3-2yz)$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} =\oint_S (3-2yz)\hat{i}\cdot dydz\hat{i} = \oint_S (3-2yz) dy dz$
Here when $y$ varies from $0$ to $1$, $z$ varies from $0$ to $-2y+2$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \int_0^{-2y+2}(3-2yz)dzdy$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1\big[ \big[3z-yz^2\big]_0^{-2y+2}dz\big]dy$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \big[3(-2y+2)-y(-2y+2)^2\big]dy$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \big[-4y^3 + 8y^2 -10y+6\big]dy$
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \big[-y^4 + \frac{8}{3}y^3 -5y^2+6y\big]_0^1 $
$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = -1+\frac{8}{3}-5+6 = \frac{8}{3}$
Hence,
$\oint_L \vec{v}\cdot \vec{dl} = \oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da}= \frac{8}{3}$
Hence, Stokes theorem is verified.
