To find the electric force experienced by a charge, we first need to determine the electric field $\mathbf{E}$ at the point $(1, 1, 1). \\$The electric field $\mathbf{E}$ is related to the electric potential $V$ by the relation:$\\ \mathbf{E} = -\nabla V \\$Given the potential $V(x, y, z) = 6x - 8xy - 8y + 6yz,$ we find the components of $\nabla V: \\$The $x$-component of $\nabla V$ is:$\\ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6x - 8xy - 8y + 6yz) \\ = 6 - 8y \\$The $y$-component of $\nabla V$ is:$\\ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6x - 8xy - 8y + 6yz) \\ = -8x - 8 + 6z \\$The $z$-component of $\nabla V$ is:$\\ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6x - 8xy - 8y + 6yz) \\ = 6y \\$Thus, the electric field $\mathbf{E}$ is:$\\ \mathbf{E} = -\left( (6 - 8y) \hat{i} + (-8x - 8 + 6z) \hat{j} + 6y \hat{k} \right) \\$Substitute $(x, y, z) = (1, 1, 1)$ into the expression for $\mathbf{E}: \\ \mathbf{E} = -\left( (6 - 8 \cdot 1) \hat{i} + (-8 \cdot 1 - 8 + 6 \cdot 1) \hat{j} + 6 \cdot 1 \hat{k} \right) \\ = -\left( (6 - 8) \hat{i} + (-8 - 8 + 6) \hat{j} + 6 \hat{k} \right) \\ = -\left( -2 \hat{i} - 10 \hat{j} + 6 \hat{k} \right) \\ = 2 \hat{i} + 10 \hat{j} - 6 \hat{k} \\$The electric force $\mathbf{F}$ on a charge $q$ is given by:$\\ \mathbf{F} = q \mathbf{E} \\$For a charge of $2 \, \mathrm{C},$ the force is:$\\ \mathbf{F} = 2 (2 \hat{i} + 10 \hat{j} - 6 \hat{k}) \\ = 4 \hat{i} + 20 \hat{j} - 12 \hat{k} \\$The magnitude of the force $|\mathbf{F}|$ is:$\\ |\mathbf{F}| = \sqrt{(4)^2 + (20)^2 + (-12)^2} \\ = \sqrt{16 + 400 + 144} \\ = \sqrt{560} \\ = \sqrt{4 \times 140} \\ = 2 \sqrt{140} \\ = 4 \sqrt{35} \\$Therefore, the electric force experienced by the charge is $4 \sqrt{35} \, \mathrm{N}. \\$This corresponds to Option 4.