To determine the electric field from the given potential function, we need to use the relationship between electric field $$\mathbf{E}$$ and electric potential $$\mathrm{V}$$. The electric field is the negative gradient of the potential:

$$\mathbf{E} = -\nabla \mathrm{V}$$

Given the potential function $$\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{xy}-\mathrm{y}+2 \mathrm{yz}$$, we need to compute the partial derivatives of $$\mathrm{V}$$ with respect to $$\mathrm{x}$$, $$\mathrm{y}$$, and $$\mathrm{z}$$.

First, calculate the partial derivative with respect to $$\mathrm{x}$$:

$$\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = \frac{\partial}{\partial \mathrm{x}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 6 \mathrm{y}$$

Next, calculate the partial derivative with respect to $$\mathrm{y}$$:

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

Finally, calculate the partial derivative with respect to $$\mathrm{z}$$:

$$\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = \frac{\partial}{\partial \mathrm{z}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 2 \mathrm{y}$$

Now, the components of the electric field are:

$$ \mathbf{E}_x = -\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = -6 \mathrm{y} $$

$$ \mathbf{E}_y = -\frac{\partial \mathrm{V}}{\partial \mathrm{y}} = -(6 \mathrm{x} - 1 + 2 \mathrm{z}) $$

$$ \mathbf{E}_z = -\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = -2 \mathrm{y} $$

We are asked to find the electric field at the point $$(1,0,1)$$.

Substitute $$\mathrm{x} = 1$$, $$\mathrm{y} = 0$$, and $$\mathrm{z} = 1$$ into the components of the electric field:

$$ \mathbf{E}_x(1,0,1) = -6(0) = 0 $$

$$ \mathbf{E}_y(1,0,1) = -(6(1) - 1 + 2(1)) = -(6 - 1 + 2) = -(7) = -7 $$

$$ \mathbf{E}_z(1,0,1) = -2(0) = 0 $$

Therefore, the electric field at the point $$(1,0,1)$$ is:

$$\mathbf{E} = 0 \mathbf{i} - 7 \mathbf{j} + 0 \mathbf{k} = -7 \mathbf{j}$$

Hence, the correct option is:

Option A: $$\mathrm{-7 j}$$

Electrostatics

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