Let an ant starts from the origin (O) and travels 2 units on negative x - axis, 3 units on positive y - axis and travels 3 units on negative z - axis to reach at point A. If `veca=hati-3hatj+2hatk and vecb` be such that the resultant of `veca and vecb` is `3hati-4hatj+hatk`, then `|vec(OA)xx(vecaxx vecb)|^(2)` is equal to

Answer

To solve the problem step by step, we will first determine the position vector of point A, then find the vector B, and finally compute the required expression.

### Step 1: Determine the position vector OA
The ant travels as follows:
- 2 units on the negative x-axis: This gives us \(-2 \hat{i}\).
- 3 units on the positive y-axis: This gives us \(+3 \hat{j}\).
- 3 units on the negative z-axis: This gives us \(-3 \hat{k}\).

Thus, the position vector \(\vec{OA}\) is:
\[ \vec{OA} = -2 \hat{i} + 3 \hat{j} - 3 \hat{k} \]

### Step 2: Given vector A
The vector \(\vec{A}\) is given as:
\[ \vec{A} = \hat{i} - 3 \hat{j} + 2 \hat{k} \]

### Step 3: Find vector B
We know the resultant of vectors \(\vec{A}\) and \(\vec{B}\) is given as:
\[ \vec{A} + \vec{B} = 3 \hat{i} - 4 \hat{j} + \hat{k} \]

To find \(\vec{B}\), we rearrange the equation:
\[ \vec{B} = (3 \hat{i} - 4 \hat{j} + \hat{k}) - \vec{A} \]
Substituting \(\vec{A}\):
\[ \vec{B} = (3 \hat{i} - 4 \hat{j} + \hat{k}) - (\hat{i} - 3 \hat{j} + 2 \hat{k}) \]
Calculating this gives:
\[ \vec{B} = (3 - 1) \hat{i} + (-4 + 3) \hat{j} + (1 - 2) \hat{k} = 2 \hat{i} - \hat{j} - \hat{k} \]

### Step 4: Calculate \(|\vec{OA} \times (\vec{A} \times \vec{B})|^2\)
We need to find \(\vec{A} \times \vec{B}\) first. The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & -1 & -1 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -3 & 2 \\ -1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 2 & -1 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} -3 & 2 \\ -1 & -1 \end{vmatrix} = (-3)(-1) - (2)(-1) = 3 + 2 = 5\)
2. \(\begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5\)
3. \(\begin{vmatrix} 1 & -3 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (-3)(2) = -1 + 6 = 5\)

Thus,
\[ \vec{A} \times \vec{B} = 5 \hat{i} + 5 \hat{j} + 5 \hat{k} = 5(\hat{i} + \hat{j} + \hat{k}) \]

### Step 5: Now calculate \(\vec{OA} \times (\vec{A} \times \vec{B})\)
Using the vector \(\vec{OA} = -2 \hat{i} + 3 \hat{j} - 3 \hat{k}\):
\[ \vec{OA} \times (5(\hat{i} + \hat{j} + \hat{k})) = 5 \vec{OA} \times (\hat{i} + \hat{j} + \hat{k}) \]
Calculating this cross product:
\[ \vec{OA} \times (\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & -3 \\ 1 & 1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 3 & -3 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & -3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 3 & -3 \\ 1 & 1 \end{vmatrix} = (3)(1) - (-3)(1) = 3 + 3 = 6\)
2. \(\begin{vmatrix} -2 & -3 \\ 1 & 1 \end{vmatrix} = (-2)(1) - (-3)(1) = -2 + 3 = 1\)
3. \(\begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} = (-2)(1) - (3)(1) = -2 - 3 = -5\)

Thus,
\[ \vec{OA} \times (\hat{i} + \hat{j} + \hat{k}) = 6 \hat{i} - 1 \hat{j} - 5 \hat{k} \]
So,
\[ \vec{OA} \times (5(\hat{i} + \hat{j} + \hat{k})) = 5(6 \hat{i} - 1 \hat{j} - 5 \hat{k}) = 30 \hat{i} - 5 \hat{j} - 25 \hat{k} \]

### Step 6: Calculate the magnitude squared
Now we need to find the magnitude squared:
\[ |\vec{OA} \times (\vec{A} \times \vec{B})|^2 = |30 \hat{i} - 5 \hat{j} - 25 \hat{k}|^2 \]
Calculating this:
\[ = 30^2 + (-5)^2 + (-25)^2 = 900 + 25 + 625 = 1550 \]

### Final Answer
Thus, the value of \(|\vec{OA} \times (\vec{A} \times \vec{B})|^2\) is:
\[ \boxed{1550} \]