(3)

The equipotential surface is always perpendicular to the electric field lines. So, the angle between the electric lines of force and the equipotential surface is 90°.

(1)

Work done is given as $W=q\Delta V$

In all the four cases the potential difference from A to B is same.

$\therefore$    In all the four cases the work done is same.

(4)

The electric field $\overrightarrow{E}$ and potential V in a region are related as $\overrightarrow{E}=-[\frac{{\displaystyle \partial V}}{{\displaystyle \partial x}}\hat{i}+\frac{{\displaystyle \partial V}}{{\displaystyle \partial y}}\hat{j}+\frac{{\displaystyle \partial V}}{{\displaystyle \partial z}}\hat{k}]$

Here, $V(x,y,z)=6xy-y+2yz$

$\therefore \overrightarrow{E}=-\left[\frac{{\displaystyle \partial }}{{\displaystyle \partial x}}\right(6xy-y+2yz)\hat{i}+\frac{{\displaystyle \partial }}{{\displaystyle \partial y}}(6xy-y+2yz)\hat{j}+\frac{{\displaystyle \partial }}{{\displaystyle \partial z}}(6xy-y+2yz\left)\hat{k}\right]$

         $=-\left[\right(6y)\hat{i}+(6x-1+2z)\hat{j}+(2y\left)\hat{k}\right]$

At point (1, 1, 0),

$\overrightarrow{E}=-\left[\right(6\left(1\right))\hat{i}+(6\left(1\right)-1+2\left(0\right))\hat{j}+(2\left(1\right)\left)\hat{k}\right]=-(6\hat{i}+5\hat{j}+2\hat{k})$

(4)

Here, $V(x,y,z)=6x-8xy-8y+6yz$

The $x,y$ and $z$ components of electric field are

$E_x=-\frac{\partial V}{\partial x}=-\frac{\partial }{\partial x}(6x-8xy-8y+6yz)=-(6-8y)=-6+8y$

$E_y=-\frac{\partial V}{\partial y}=-\frac{\partial }{\partial y}(6x-8xy-8y+6yz)=-(-8x-8+6z)=8x+8-6z$

$E_z=-\frac{\partial V}{\partial z}=-\frac{\partial }{\partial z}(6x-8xy-8y+6yz)=-6y$

$\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$

$=(-6+8y)\hat{i}+(8x+8-6z)\hat{j}-6y\hat{k}$

At point (1, 1, 1)

$\overrightarrow{E}=(-6+8)\hat{i}+(8+8-6)\hat{j}-6\hat{k}=2\hat{i}+10\hat{j}-6\hat{k}$

The magnitude of electric field $\overrightarrow{E}$ is

$\overrightarrow{E}=\sqrt{E_x^2+E_y^2+E_z^2}=\sqrt{2^2+{10}^2+{(-6)}^2}$

$=\sqrt{140}=2\sqrt{35}$$\text{N}$${\mathrm{C}}^{-1}$

Electric force experienced by the charge
$F=qE=2 \text{C}\times 2\sqrt{35} {\text{N C}}^{-1}=4\sqrt{35}$$\text{N}$