Two unlike force of equal magnitudes `hat(j)+2hat(k)` and `-hat(j)-2hat(k)` are acting at the points whose position vectors are given by `hat(i)+hat(j)+hat(k)` and `hat(i)+2hat(j)+3hat(k)` respectively. Find the moment of the couple formed by these forces.

Answer

To find the moment of the couple formed by the two forces, we can follow these steps:

### Step 1: Identify the Forces and Position Vectors
The forces acting on the points are:
- **Force F1** at point A: \( \hat{j} + 2\hat{k} \)
- **Force F2** at point B: \( -\hat{j} - 2\hat{k} \)

The position vectors of the points are:
- Point A: \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
- Point B: \( \vec{B} = \hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Find the Position Vector from A to B
To find the vector \( \vec{BA} \):
\[ \vec{BA} = \vec{A} - \vec{B} \]
Substituting the values:
\[ \vec{BA} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} + \hat{j} + \hat{k} - \hat{i} - 2\hat{j} - 3\hat{k} \]
Simplifying this gives:
\[ \vec{BA} = 0\hat{i} - \hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k} \]

### Step 3: Calculate the Moment of the Couple
The moment \( \vec{M} \) of the couple is given by the cross product of the position vector \( \vec{BA} \) and one of the forces (we can use \( F1 \)):
\[ \vec{M} = \vec{BA} \times \vec{F1} \]
Substituting the vectors:
\[ \vec{M} = (-\hat{j} - 2\hat{k}) \times (\hat{j} + 2\hat{k}) \]

### Step 4: Compute the Cross Product
Using the determinant method for the cross product:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & -2 \\ 0 & 1 & 2 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{M} = \hat{i} \begin{vmatrix} -1 & -2 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -2 \\ 0 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & -1 \\ 0 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. For \( \hat{i} \):
\[ (-1)(2) - (-2)(1) = -2 + 2 = 0 \]
2. For \( \hat{j} \):
\[ 0 - 0 = 0 \]
3. For \( \hat{k} \):
\[ 0 - 0 = 0 \]

Thus:
\[ \vec{M} = 0\hat{i} - 0\hat{j} + 0\hat{k} = \vec{0} \]

### Conclusion
The moment of the couple formed by these forces is:
\[ \vec{M} = \vec{0} \]