Forces `2hat(i)+hat(j), 2hat(i)-3hat(j)+6hat(k) and hat(i)+2hat(j)-hat(k)` act at a point P, with position vector `4hat(i)-3hat(j)-hat(k)`. Find the moment of the resultant of these force about the point Q whose position vector is `6hat(i)+hat(j)-3hat(k)`.

Answer

To solve the problem, we need to find the moment of the resultant of the given forces about point Q. Here’s a step-by-step solution:

### Step 1: Identify the Forces
The forces acting at point P are given as:
- \( \mathbf{F_1} = 2\hat{i} + \hat{j} \)
- \( \mathbf{F_2} = 2\hat{i} - 3\hat{j} + 6\hat{k} \)
- \( \mathbf{F_3} = \hat{i} + 2\hat{j} - \hat{k} \)

### Step 2: Calculate the Resultant Force
To find the resultant force \( \mathbf{F} \), we add the three forces together:
\[ \mathbf{F} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} \]
Calculating each component:
- **i-component**: \( 2 + 2 + 1 = 5 \)
- **j-component**: \( 1 - 3 + 2 = 0 \)
- **k-component**: \( 0 + 6 - 1 = 5 \)

Thus, the resultant force is:
\[ \mathbf{F} = 5\hat{i} + 0\hat{j} + 5\hat{k} = 5\hat{i} + 5\hat{k} \]

### Step 3: Find the Position Vectors
The position vector of point P is given as:
\[ \mathbf{P} = 4\hat{i} - 3\hat{j} - \hat{k} \]
The position vector of point Q is given as:
\[ \mathbf{Q} = 6\hat{i} + \hat{j} - 3\hat{k} \]

### Step 4: Calculate the Vector \( \mathbf{PQ} \)
The vector from point Q to point P is given by:
\[ \mathbf{PQ} = \mathbf{P} - \mathbf{Q} \]
Calculating this:
\[ \mathbf{PQ} = (4\hat{i} - 3\hat{j} - \hat{k}) - (6\hat{i} + \hat{j} - 3\hat{k}) = (4 - 6)\hat{i} + (-3 - 1)\hat{j} + (-1 + 3)\hat{k} \]
This simplifies to:
\[ \mathbf{PQ} = -2\hat{i} - 4\hat{j} + 2\hat{k} \]

### Step 5: Calculate the Moment of the Resultant Force
The moment \( \mathbf{M} \) of the resultant force about point Q is given by the cross product:
\[ \mathbf{M} = \mathbf{PQ} \times \mathbf{F} \]
Substituting the vectors:
\[ \mathbf{M} = (-2\hat{i} - 4\hat{j} + 2\hat{k}) \times (5\hat{i} + 0\hat{j} + 5\hat{k}) \]
Setting up the determinant:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -4 & 2 \\ 5 & 0 & 5 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} -4 & 2 \\ 0 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 2 \\ 5 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & -4 \\ 5 & 0 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -4 & 2 \\ 0 & 5 \end{vmatrix} = (-4)(5) - (0)(2) = -20 \)
2. \( \begin{vmatrix} -2 & 2 \\ 5 & 5 \end{vmatrix} = (-2)(5) - (2)(5) = -10 - 10 = -20 \)
3. \( \begin{vmatrix} -2 & -4 \\ 5 & 0 \end{vmatrix} = (-2)(0) - (-4)(5) = 0 + 20 = 20 \)

Putting it all together:
\[ \mathbf{M} = -20\hat{i} + 20\hat{j} + 20\hat{k} \]

### Final Answer
Thus, the moment of the resultant force about point Q is:
\[ \mathbf{M} = -20\hat{i} + 20\hat{j} + 20\hat{k} \]