A force $$\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$$ is applied on an intersection point of $$x = 2$$ plane and $$x$$-axis. The magnitude of torque of this force about a point $$(2, 3, 4)$$ is ________. (Round off to the Nearest Integer)

The intersection of the $$x = 2$$ plane and the $$x$$-axis is the point $$(2, 0, 0)$$. The force is $$\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$$ applied at point $$A = (2, 0, 0)$$.

We need the torque about point $$B = (2, 3, 4)$$. The position vector from $$B$$ to $$A$$ is $$\vec{r} = A - B = (2-2)\hat{i} + (0-3)\hat{j} + (0-4)\hat{k} = -3\hat{j} - 4\hat{k}$$.

The torque is $$\vec{\tau} = \vec{r} \times \vec{F}$$:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -4 \\ 4 & 3 & 4 \end{vmatrix}$$

$$= \hat{i}[(-3)(4) - (-4)(3)] - \hat{j}[(0)(4) - (-4)(4)] + \hat{k}[(0)(3) - (-3)(4)]$$

$$= \hat{i}[-12 + 12] - \hat{j}[0 + 16] + \hat{k}[0 + 12]$$

$$= 0\hat{i} - 16\hat{j} + 12\hat{k}$$

The magnitude is $$|\vec{\tau}| = \sqrt{0^2 + 16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$.

The magnitude of torque is $$\boxed{20}$$.