Find the torque of a force `F=a(hati+2hatj+3hatk)` `N` about a point O. The position vector of point of application of force about `O` is `r=(2hati+3hatj-hatk)` `m`.

Explanation

To find the torque \( \tau \) of a force \( \mathbf{F} = a(\hat{i} + 2\hat{j} + 3\hat{k}) \) about a point \( O \), given the position vector \( \mathbf{r} = (2\hat{i} + 3\hat{j} - \hat{k}) \), we can follow these steps:

### Step 1: Write down the given vectors
We have:
- Force vector:
\[ \mathbf{F} = a(\hat{i} + 2\hat{j} + 3\hat{k}) = a\hat{i} + 2a\hat{j} + 3a\hat{k} \]
...