Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $$\vec{F} = -P\hat{k}$$.

To find the torque, we first find the position vector of point A with respect to point B:

$$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 2)\hat{i} + (0 - (-3))\hat{j} + (0 - 0)\hat{k} = -2\hat{i} + 3\hat{j}$$

To calculate the cross product, we can use the determinant method with a 3x3 matrix:

$$\vec{\tau} = \vec{r}_{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ 0 & 0 & -P \\ \end{vmatrix}$$

Now, we will calculate the cross product components by expanding the determinant along the first row:

1. $$\tau_i = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -P \\ \end{vmatrix} = \hat{i}((3)(-P) - (0)(0)) = -3P\hat{i}$$

2. $$\tau_j = -\hat{j} \begin{vmatrix} -2 & 0 \\ 0 & -P \\ \end{vmatrix} = -\hat{j}((-2)(-P) - (0)(0)) = -2P\hat{j}$$

(Notice the negative sign in front of the $$\hat{j}$$ term, as it comes from the expansion of the determinant.)

3. $$\tau_k = \hat{k} \begin{vmatrix} -2 & 3 \\ 0 & 0 \\ \end{vmatrix} = \hat{k}((-2)(0) - (3)(0)) = 0\hat{k}$$

Now, combine the components to get the torque vector:

$$\vec{\tau} = -3P\hat{i} - 2P\hat{j} + 0\hat{k} = -3P\hat{i} - 2P\hat{j}$$

Comparing this to the given torque vector $$\vec{\tau} = P(a\hat{i} + b\hat{j})$$, we find that:

$$a = -3$$

$$b = -2$$

Thus, the ratio $$\frac{a}{b} = \frac{-3}{-2} = \frac{x}{2}$$.

Therefore, $$x = 3$$.