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, let's break it down:

### Step 1: Determine the position vector OA
The ant starts from the origin (O) and travels:
- 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 \( \vec{OA} \) can be expressed as:
\[ \vec{OA} = -2\hat{i} + 3\hat{j} - 3\hat{k} \]

### Step 2: Identify 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 by:
\[ \vec{A} + \vec{B} = 3\hat{i} - 4\hat{j} + \hat{k} \]
To find \( \vec{B} \), we rearrange this 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 \( \vec{OA} \cdot \vec{B} \)
Now, we compute the dot product \( \vec{OA} \cdot \vec{B} \):
\[ \vec{OA} \cdot \vec{B} = (-2\hat{i} + 3\hat{j} - 3\hat{k}) \cdot (2\hat{i} - \hat{j} - \hat{k}) \]
Calculating this gives:
\[ = (-2)(2) + (3)(-1) + (-3)(-1) = -4 - 3 + 3 = -4 \]

### Step 5: Calculate \( \vec{OA} \cdot \vec{A} \)
Next, we calculate \( \vec{OA} \cdot \vec{A} \):
\[ \vec{OA} \cdot \vec{A} = (-2\hat{i} + 3\hat{j} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 2\hat{k}) \]
Calculating this gives:
\[ = (-2)(1) + (3)(-3) + (-3)(2) = -2 - 9 - 6 = -17 \]

### Step 6: Calculate \( \vec{OA} \times (\vec{A} \times \vec{B}) \)
Using the formula \( \vec{OA} \times (\vec{A} \times \vec{B}) = \vec{OA} \cdot \vec{B} \cdot \vec{A} - \vec{OA} \cdot \vec{A} \cdot \vec{B} \):
\[ \vec{OA} \times (\vec{A} \times \vec{B}) = -4\vec{A} + 17\vec{B} \]
Substituting the values of \( \vec{A} \) and \( \vec{B} \):
\[ = -4(\hat{i} - 3\hat{j} + 2\hat{k}) + 17(2\hat{i} - \hat{j} - \hat{k}) \]
Calculating this gives:
\[ = (-4\hat{i} + 12\hat{j} - 8\hat{k}) + (34\hat{i} - 17\hat{j} - 17\hat{k}) \]
Combining like terms:
\[ = (30\hat{i} - 5\hat{j} - 25\hat{k}) \]

### Step 7: Calculate the magnitude squared
Now, we calculate the magnitude squared of the resultant vector:
\[ |\vec{OA} \times (\vec{A} \times \vec{B})|^2 = (30)^2 + (-5)^2 + (-25)^2 \]
Calculating this gives:
\[ = 900 + 25 + 625 = 1550 \]

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