If `|{:(veca,vecb,vecc),(veca.veca,veca.vecb,veca.vecc),(veca.vecc,vecb.vecc,veca.vecc)|` where `veca, vecb,vecc` are coplanar then:

Options

A. `vec(Delta) = vec0`
B. `vecDelta = veca + vecb + vecc`
C. `vecDelta` = any non-zero vectors
D. None of these

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{c} & \vec{c} \cdot \vec{c} & \vec{a} \cdot \vec{c} \end{vmatrix} \]

Given that \(\vec{a}, \vec{b}, \vec{c}\) are coplanar, we need to show that the determinant equals zero.

### Step 1: Set Up the Vectors
Assume:
- \(\vec{a} = \hat{i}\)
- \(\vec{b} = 2\hat{i}\)
- \(\vec{c} = 3\hat{i}\)

### Step 2: Calculate the Dot Products
Now, we compute the dot products:
- \(\vec{a} \cdot \vec{a} = \hat{i} \cdot \hat{i} = 1\)
- \(\vec{a} \cdot \vec{b} = \hat{i} \cdot (2\hat{i}) = 2\)
- \(\vec{a} \cdot \vec{c} = \hat{i} \cdot (3\hat{i}) = 3\)
- \(\vec{b} \cdot \vec{c} = (2\hat{i}) \cdot (3\hat{i}) = 6\)
- \(\vec{c} \cdot \vec{c} = (3\hat{i}) \cdot (3\hat{i}) = 9\)

### Step 3: Substitute into the Determinant
Now substitute these values into the determinant:

\[ \Delta = \begin{vmatrix} \hat{i} & 2\hat{i} & 3\hat{i} \\ 1 & 2 & 3 \\ 6 & 9 & 3 \end{vmatrix} \]

### Step 4: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ \Delta = \hat{i} \begin{vmatrix} 2 & 3 \\ 9 & 3 \end{vmatrix} - 2\hat{i} \begin{vmatrix} 1 & 3 \\ 6 & 3 \end{vmatrix} + 3\hat{i} \begin{vmatrix} 1 & 2 \\ 6 & 9 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 2 & 3 \\ 9 & 3 \end{vmatrix} = (2)(3) - (3)(9) = 6 - 27 = -21\)

2. \(\begin{vmatrix} 1 & 3 \\ 6 & 3 \end{vmatrix} = (1)(3) - (3)(6) = 3 - 18 = -15\)

3. \(\begin{vmatrix} 1 & 2 \\ 6 & 9 \end{vmatrix} = (1)(9) - (2)(6) = 9 - 12 = -3\)

Now substituting back into the determinant:

\[ \Delta = \hat{i}(-21) - 2\hat{i}(-15) + 3\hat{i}(-3) \]

\[ = -21\hat{i} + 30\hat{i} - 9\hat{i} \]

\[ = (-21 + 30 - 9)\hat{i} = 0\hat{i} \]

### Conclusion
Thus, \(\Delta = 0\).

### Final Answer
The value of \(\Delta\) is \(0\).

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