1. **State the problem:** We need to find the missing coordinate $x$ of point $A(x, 2)$ such that the moment $M_0$ created by the force $\mathbf{F} = 3\mathbf{i} - 4\mathbf{j}$ about the origin is $14\mathbf{k}$. 2. **Recall the formula for moment:** The moment $\mathbf{M}_0$ about the origin due to a force $\mathbf{F}$ applied at point $\mathbf{r}$ is given by the cross product: $$\mathbf{M}_0 = \mathbf{r} \times \mathbf{F}$$ where $\mathbf{r} = x\mathbf{i} + 2\mathbf{j}$. 3. **Calculate the cross product:** $$\mathbf{M}_0 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x & 2 & 0 \\ 3 & -4 & 0 \end{vmatrix} = (2 \cdot 0 - 0 \cdot (-4))\mathbf{i} - (x \cdot 0 - 0 \cdot 3)\mathbf{j} + (x \cdot (-4) - 2 \cdot 3)\mathbf{k}$$ $$= 0\mathbf{i} - 0\mathbf{j} + (-4x - 6)\mathbf{k} = (-4x - 6)\mathbf{k}$$ 4. **Set the moment equal to given moment:** $$-4x - 6 = 14$$ 5. **Solve for $x$:** $$-4x = 14 + 6$$ $$-4x = 20$$ $$x = \frac{\cancel{-4}x}{\cancel{-4}} = \frac{20}{-4} = -5$$ 6. **Answer:** The missing coordinate $x$ is $-5$. **Final answer:** $\boxed{-5}$