1. Given data

   Force: $\vec F = \hat i + 2\hat j + 3\hat k$

   Point of application of force: $\vec r_1 = 4\hat i + 3\hat j - \hat k$

   Point about which torque is to be calculated: $\vec r_0 = \hat i + 2\hat j + \hat k$

2. Position vector from torque point to point of application

   $\vec r = \vec r_1 - \vec r_0$ $= (4-1)\hat i + (3-2)\hat j + (-1-1)\hat k$ $= 3\hat i + \hat j - 2\hat k$

3. Torque formula

   Torque about the given point is $\vec \tau = \vec r \times \vec F$

   So,

   \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & -2 \\ 1 & 2 & 3 \end{vmatrix}$$

4. Evaluate the cross product

   - \hat j\big(3\cdot 3 - (-2)\cdot 1\big) + \hat k\big(3\cdot 2 - 1\cdot 1\big)$$ $$\vec \tau = \hat i(3+4) - \hat j(9+2) + \hat k(6-1)$$ $$\vec \tau = 7\hat i - 11\hat j + 5\hat k$$

5. Magnitude of torque

   $|\vec \tau| = \sqrt{7^2 + (-11)^2 + 5^2}$ $= \sqrt{49 + 121 + 25}$ $= \sqrt{195}$

   Comparing with $\sqrt{x}$, $x = 195$

6. Comparison with stored answer

   Stored correct answer = $195$

   Our derived answer also is $195$.

Final Answer: $\boxed{195}$