Answer
The work $W$ against ${\bf{F}}$ required to move a particle of charge $q = 0.01$ $C$ from $\left( {1, - 5,0} \right)$ to $\left( {3,4,4} \right)$ is
$W \approx 54.79$ joules
Work Step by Step
We have an electric field (in newtons per coulomb) given by
${\bf{F}}\left( {x,y,z} \right) = \dfrac{{kp}}{{{r^5}}}\left( {3xz,3yz,2{z^2} - {x^2} - {y^2}} \right)$
where $r = {\left( {{x^2} + {y^2} + {z^2}} \right)^{1/2}}$ with distance in meters.
Since the electric field is conservative, let the potential energy for the electric force $q{\bf{F}}$ be $V\left( {x,y,z} \right)$ such that $q{\bf{F}} = - \nabla V = - \left( {\dfrac{{\partial V}}{{\partial x}},\dfrac{{\partial V}}{{\partial y}},\dfrac{{\partial V}}{{\partial z}}} \right)$. So,
$\nabla V = - \dfrac{{qkp}}{{{r^5}}}\left( {3xz,3yz,2{z^2} - {x^2} - {y^2}} \right)$
a. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives
$V\left( {x,y,z} \right) = - qkp\smallint \dfrac{{3xz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{5/2}}}}{\rm{d}}x = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + f\left( {y,z} \right)$
b. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives
$V\left( {x,y,z} \right) = - qkp\smallint \dfrac{{3yz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{5/2}}}}{\rm{d}}y = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + g\left( {x,z} \right)$
c. taking the integral of $\dfrac{{\partial V}}{{\partial z}}$ with respect to $z$ gives
$V\left( {x,y,z} \right) = - qkp\smallint \dfrac{{2{z^2} - {x^2} - {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{5/2}}}}{\rm{d}}z = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + h\left( {x,y} \right)$
Since the three ways of expressing $V\left( {x,y,z} \right)$ must be equal, we get
$qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + f\left( {y,z} \right) = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + g\left( {x,z} \right) = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + h\left( {x,y} \right)$
Thus, we conclude that the potential energy is $V\left( {x,y,z} \right) = qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + C$, where $C$ is a constant.
So the work $W$ against ${\bf{F}}$ required to move a particle of charge $q = 0.01$ $C$ from $\left( {1, - 5,0} \right)$ to $\left( {3,4,4} \right)$ is
$W = \left[ {qkp\dfrac{z}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + C} \right]_{\left( {1, - 5,0} \right)}^{\left( {3,4,4} \right)}$
$W = 0.01\cdot8.99 \times {10^9}\cdot4 \times {10^{ - 5}}\left( {\dfrac{4}{{{{41}^{3/2}}}} - 0} \right)$
$W \approx 54.79$ joules
