A three coulomb charge moves from the point \((0, -2, -5)\) to the point \((5, 1, 2)\) in an electric field expressed as \(\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}\) N/C. The work done in moving the charge is _______ J.

The force experienced by the charge in the electric field is given by \(\vec{F} = q\vec{E}\).

Given \(q = 3\) C and \(\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}\) N/C, the force is:
\(\vec{F} = 3(2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) = 6x\hat{i} + 9y^2\hat{j} + 12\hat{k}\) N

The work done by the electric field in moving the charge is:
\(W = \int \vec{F} \cdot d\vec{r} = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz\)

Substituting the limits from the initial point \((0, -2, -5)\) to the final point \((5, 1, 2)\):
\(W = \int_{0}^{5} 6x dx + \int_{-2}^{1} 9y^2 dy + \int_{-5}^{2} 12 dz\)

Evaluating each integral separately:
\(\int_{0}^{5} 6x dx = \left[ 3x^2 \right]_{0}^{5} = 3(5^2 - 0^2) = 75\) J

\(\int_{-2}^{1} 9y^2 dy = \left[ 3y^3 \right]_{-2}^{1} = 3(1^3 - (-2)^3) = 3(1 - (-8)) = 3(9) = 27\) J

\(\int_{-5}^{2} 12 dz = \left[ 12z \right]_{-5}^{2} = 12(2 - (-5)) = 12(7) = 84\) J

Total work done \(W = 75 + 27 + 84 = 186\) J.

Answer: \(186\)