The vectors
`veca=xhati+(x+1)hatj+(x+2)hatk`,
`vecb=(x+3)hati+(x+4)hatj+(x+5)hatk`
and `vecc=(x+6)hati+(x+7)hatj+(x+8)hatk` are coplanar for

Options

A. all values of `x`
B. `x lt 0` only
C. `x gt0` only
D. None of these

Explanation

To determine the values of \( x \) for which the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar, we can use the scalar triple product. The vectors are given as follows:

\[ \vec{a} = x \hat{i} + (x + 1) \hat{j} + (x + 2) \hat{k} \]
\[ \vec{b} = (x + 3) \hat{i} + (x + 4) \hat{j} + (x + 5) \hat{k} \]
\[ \vec{c} = (x + 6) \hat{i} + (x + 7) \hat{j} + (x + 8) \hat{k} \]

### Step 1: Set up the determinant for the scalar triple product

The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar if the scalar triple product is zero, which can be represented as the determinant of the matrix formed by the coefficients of the vectors:

\[ \begin{vmatrix} x & x + 1 & x + 2 \\ x + 3 & x + 4 & x + 5 \\ x + 6 & x + 7 & x + 8 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant

We can expand the determinant:

\[ D = \begin{vmatrix} x & x + 1 & x + 2 \\ x + 3 & x + 4 & x + 5 \\ x + 6 & x + 7 & x + 8 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = x \begin{vmatrix} x + 4 & x + 5 \\ x + 7 & x + 8 \end{vmatrix} - (x + 1) \begin{vmatrix} x + 3 & x + 5 \\ x + 6 & x + 8 \end{vmatrix} + (x + 2) \begin{vmatrix} x + 3 & x + 4 \\ x + 6 & x + 7 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

Calculating each of the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} x + 4 & x + 5 \\ x + 7 & x + 8 \end{vmatrix} = (x + 4)(x + 8) - (x + 5)(x + 7) = x^2 + 12x + 32 - (x^2 + 12x + 35) = -3 \]

2. For the second determinant:
\[ \begin{vmatrix} x + 3 & x + 5 \\ x + 6 & x + 8 \end{vmatrix} = (x + 3)(x + 8) - (x + 5)(x + 6) = x^2 + 11x + 24 - (x^2 + 11x + 30) = -6 \]

3. For the third determinant:
\[ \begin{vmatrix} x + 3 & x + 4 \\ x + 6 & x + 7 \end{vmatrix} = (x + 3)(x + 7) - (x + 4)(x + 6) = x^2 + 10x + 21 - (x^2 + 10x + 24) = -3 \]

### Step 4: Substitute back into the determinant equation

Now substituting back into the determinant equation:

\[ D = x(-3) - (x + 1)(-6) + (x + 2)(-3) \]

Expanding this:

\[ D = -3x + 6(x + 1) - 3(x + 2) \]
\[ = -3x + 6x + 6 - 3x - 6 \]
\[ = 0 \]

### Step 5: Conclusion

Since \(D = 0\) for all values of \(x\), we conclude that the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar for all values of \(x\).

### Final Answer

The vectors are coplanar for all values of \(x\).

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