If the vector `ai+j+k,i+bj+k` and `i+j+ck` are coplanar, then :

Options

A. `abc = -1`
B. `a+b+c=0`
C. `a+b+c =abc +2`
D. `ab+bc+ca=0`

Answer

To determine the condition under which the vectors \( \vec{a} = ai + j + k \), \( \vec{b} = i + bj + k \), and \( \vec{c} = i + j + ck \) are coplanar, we will use the concept of the scalar triple product. The scalar triple product of three vectors \( \vec{p}, \vec{q}, \vec{r} \) is given by the determinant of a matrix formed by the vectors. If the scalar triple product is zero, then the vectors are coplanar.

### Step-by-step Solution:

1. **Write the vectors:**
\[ \vec{p} = ai + j + k \]
\[ \vec{q} = i + bj + k \]
\[ \vec{r} = i + j + ck \]

2. **Set up the scalar triple product:**
The scalar triple product can be expressed as:
\[ \vec{p} \cdot (\vec{q} \times \vec{r}) = 0 \]
This can be represented as the determinant:
\[ \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
We will expand the determinant:
\[ = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ = a(bc - 1) - (c - 1) + (1 - b) \]
\[ = abc - a - c + 1 + 1 - b \]
\[ = abc - a - b - c + 2 \]

4. **Set the determinant equal to zero:**
For the vectors to be coplanar:
\[ abc - a - b - c + 2 = 0 \]

5. **Rearranging the equation:**
\[ abc - a - b - c + 2 = 0 \implies abc = a + b + c - 2 \]

### Final Result:
Thus, the condition for the vectors to be coplanar is:
\[ abc = a + b + c - 2 \]