The set of values of 'm' for which the vectors `vec(a) = mhati + (m + 1)hatj +(m + 8)hatk`
`vec(b) = (m + 3)hati + (m+4)hatj + (m +5) hatk` and `vec(c) = (m + 6)hati + (m + 7)hatj + (m + 8) hatk` are non - coplanar is

Options

A. R
B. `R - {1}`
C. `R - {1,2}`
D. `phi`

Answer

To determine the set of values of 'm' for which the vectors \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, we need to find when the determinant of the matrix formed by these vectors is not equal to zero.

### Step-by-step Solution:

1. **Define the Vectors:**
\[ \vec{a} = m \hat{i} + (m + 1) \hat{j} + (m + 8) \hat{k} \]
\[ \vec{b} = (m + 3) \hat{i} + (m + 4) \hat{j} + (m + 5) \hat{k} \]
\[ \vec{c} = (m + 6) \hat{i} + (m + 7) \hat{j} + (m + 8) \hat{k} \]

2. **Set Up the Matrix:**
The vectors can be represented in matrix form as:
\[ \begin{vmatrix} m & m + 1 & m + 8 \\ m + 3 & m + 4 & m + 5 \\ m + 6 & m + 7 & m + 8 \end{vmatrix} \]

3. **Calculate the Determinant:**
We will calculate the determinant of the above matrix. To simplify the calculation, we can perform row operations.

- First, perform the operation \(R_3 \rightarrow R_3 - R_1\):
\[ R_3 = (m + 6 - m) \hat{i} + (m + 7 - (m + 1)) \hat{j} + (m + 8 - (m + 8)) \hat{k} = 6 \hat{i} + 6 \hat{j} + 0 \hat{k} \]
The matrix now looks like:
\[ \begin{vmatrix} m & m + 1 & m + 8 \\ m + 3 & m + 4 & m + 5 \\ 6 & 6 & 0 \end{vmatrix} \]

- Next, perform the operation \(R_2 \rightarrow R_2 - R_1\):
\[ R_2 = ((m + 3) - m) \hat{i} + ((m + 4) - (m + 1)) \hat{j} + ((m + 5) - (m + 8)) \hat{k} = 3 \hat{i} + 3 \hat{j} - 3 \hat{k} \]
The matrix now looks like:
\[ \begin{vmatrix} m & m + 1 & m + 8 \\ 3 & 3 & -3 \\ 6 & 6 & 0 \end{vmatrix} \]

4. **Factor Out Common Terms:**
We can factor out common terms from the rows:
\[ 3 \cdot 2 \cdot \begin{vmatrix} m & m + 1 & m + 8 \\ 1 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} \]

5. **Calculate the Simplified Determinant:**
Now, we calculate the determinant:
\[ = 3 \cdot 2 \cdot \left( m \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} - (m + 1) \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} + (m + 8) \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \right) \]
The determinant simplifies to:
\[ = 3 \cdot 2 \cdot (m - (m + 1) + (m + 8) \cdot 0) = 3 \cdot 2 \cdot (-1) = -6 \]

6. **Conclusion:**
Since the determinant is a constant (-6) and not equal to zero for any value of \(m\), the vectors \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar for all values of \(m\).

### Final Answer:
The set of values of \(m\) for which the vectors are non-coplanar is:
\[ \text{All real numbers } \mathbb{R} \]