The electric field $\mathbf{E}$ is related to the electric potential $V$ by the negative gradient relation: $\mathbf{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$ [1].

Given $V = 6xy - y + 2yz$, we calculate the partial derivatives:

1. $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
2. $\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
3. $\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y$

Substitute the coordinates of the point $(1, 1, 0)$ where $x=1, y=1, z=0$:

1. $E_x = - (6(1)) = -6$
2. $E_y = - (6(1) - 1 + 2(0)) = -(6 - 1) = -5$
3. $E_z = - (2(1)) = -2$

Thus, $\mathbf{E} = -6\hat{i} - 5\hat{j} - 2\hat{k} = -(6\hat{i} + 5\hat{j} + 2\hat{k})$. (Note: The provided options appear to list the vector with positive signs, likely due to a typo in the source or OCR, but the magnitude coefficients correspond to Option B).