The vectors `x hati + (x+1)hatj + (x+2)hatk, (x+3)hati+ (x+4)hatj + (x+5)hatk and (x+6)hati + (x+7)hatj+ (x+8)hatk` are coplanar if x is equal to a. 1 b. -3 c. 4 d. 0

Options

A. 1
B. `-3`
C. `4`
D. 0

Explanation

To determine the value of \( x \) for which the given vectors are coplanar, we need to set up the problem using the concept of determinants. The vectors are:

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

The vectors are coplanar if the determinant of the matrix formed by these vectors is equal to zero.

### Step 1: Set up the determinant
We can form a matrix with these vectors as rows:

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

### Step 2: Calculate the determinant
Using the determinant formula for a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

\[ \text{Det} = x \left( (x+4)(x+8) - (x+5)(x+7) \right) - (x+1) \left( (x+3)(x+8) - (x+5)(x+6) \right) + (x+2) \left( (x+3)(x+7) - (x+4)(x+6) \right) \]

### Step 3: Simplify the determinant
Calculating each part:

1. For \( x \):
\[ (x+4)(x+8) - (x+5)(x+7) = (x^2 + 12x + 32) - (x^2 + 12x + 35) = -3 \]

2. For \( (x+1) \):
\[ (x+3)(x+8) - (x+5)(x+6) = (x^2 + 11x + 24) - (x^2 + 11x + 30) = -6 \]

3. For \( (x+2) \):
\[ (x+3)(x+7) - (x+4)(x+6) = (x^2 + 10x + 21) - (x^2 + 10x + 24) = -3 \]

Now substituting back into the determinant:

\[ \text{Det} = x(-3) - (x+1)(-6) + (x+2)(-3) \]

### Step 4: Expand and simplify
\[ \text{Det} = -3x + 6 + 6x - 3x - 6 = 0 \]
\[ \text{Det} = 0 \]

### Step 5: Solve for \( x \)
Setting the determinant equal to zero gives:

\[ 0 = 0 \]

This means the determinant is always zero, indicating that the vectors are always coplanar for any value of \( x \).

### Conclusion
Thus, the vectors are coplanar for all values of \( x \), which means all options (1, -3, 4, 0) are correct.