D

To find the torque \( \tau \) produced by the force \( \vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k} \) about a point \( (2, 3, 4) \), we use the formula:
\( \tau = \vec{r} \times \vec{F} \)
where \( \vec{r} \) is the position vector from the point about which we are calculating the torque to the point where the force is applied.
Step 1: Determine \( \vec{r} \). If the force is applied at the origin \( (0, 0, 0) \), then \( \vec{r} = (0 - 2)\hat{i} + (0 - 3)\hat{j} + (0 - 4)\hat{k} = -2\hat{i} - 3\hat{j} - 4\hat{k} \).
Step 2: Calculate the cross product:
\( \tau = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -4 \\ 4 & 3 & 4 \end{vmatrix}
= \hat{i}(\text{-3} \cdot 4 - \text{-4} \cdot 3) - \hat{j}(\text{-2} \cdot 4 - \text{-4} \cdot 4) + \hat{k}(\text{-2} \cdot 3 - \text{-3} \cdot 4)
= \hat{i}(-12 + 12) - \hat{j}(-8 + 16) + \hat{k}(-6 + 12)
= 0\hat{i} - 8\hat{j} + 6\hat{k}
\tau = \sqrt{0^2 + (-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10.
Therefore, the magnitude of torque is \( 10 \) when rounded to the nearest integer, which is option D.