If `veca=hati+hatj+hatk, vecb=hati-hatj+hatk, vec c=hati+2hatj-hatk`, then the value of `|(veca*veca,veca*vecb, veca*vecc),(vecb*veca, vecb *vecb,vecb*vecc),(vec c*veca, vec c*vec b,vec c*vec c)|` is equal to :

Options

A. 2
B. 4
C. 16
D. 64

Answer

To solve the given problem, we need to calculate the determinant of the matrix formed by the dot products of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Let's break this down step by step.

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{c} = \hat{i} + 2\hat{j} - \hat{k} \]

### Step 2: Calculate the dot products
1. **Calculate \(\vec{a} \cdot \vec{a}\)**:
\[ \vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3 \]

2. **Calculate \(\vec{b} \cdot \vec{b}\)**:
\[ \vec{b} \cdot \vec{b} = 1^2 + (-1)^2 + 1^2 = 3 \]

3. **Calculate \(\vec{c} \cdot \vec{c}\)**:
\[ \vec{c} \cdot \vec{c} = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6 \]

4. **Calculate \(\vec{a} \cdot \vec{b}\)**:
\[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1 \]

5. **Calculate \(\vec{a} \cdot \vec{c}\)**:
\[ \vec{a} \cdot \vec{c} = (1)(1) + (1)(2) + (1)(-1) = 1 + 2 - 1 = 2 \]

6. **Calculate \(\vec{b} \cdot \vec{c}\)**:
\[ \vec{b} \cdot \vec{c} = (1)(1) + (-1)(2) + (1)(-1) = 1 - 2 - 1 = -2 \]

### Step 3: Form the determinant matrix
Now we can form the matrix using these dot products:
\[ \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix} = \begin{vmatrix} 3 & 1 & 2 \\ 1 & 3 & -2 \\ 2 & -2 & 6 \end{vmatrix} \]

### Step 4: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]
For our matrix:
- \(a = 3\), \(b = 1\), \(c = 2\)
- \(d = 1\), \(e = 3\), \(f = -2\)
- \(g = 2\), \(h = -2\), \(i = 6\)

Calculating:
\[ \text{Det} = 3(3 \cdot 6 - (-2)(-2)) - 1(1 \cdot 6 - (-2)(2)) + 2(1 \cdot (-2) - 3 \cdot 2) \]
\[ = 3(18 - 4) - 1(6 + 4) + 2(-2 - 6) \]
\[ = 3(14) - 1(10) + 2(-8) \]
\[ = 42 - 10 - 16 \]
\[ = 42 - 26 = 16 \]

### Final Answer
The value of the determinant is:
\[ \boxed{16} \]