If the vectors `veca=hati+ahatj+a^(2)hatk, vecb=hati+bhatj+b^(2)hatk, vecc=hati+chatj+c^(2)hatk` are three non-coplanar vectors and `|(a, a^(2), 1+a^(3)),(b,b^(2),1+b^(3)),(c,c^(2),1+c^(3))|=0` , then the value of `abc` is

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we need to find the value of \( abc \) given the condition that the determinant

\[ \left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0 \]

### Step 1: Write the determinant
We start with the determinant:

\[ D = \left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| \]

### Step 2: Split the determinant
Using the property of determinants, we can express the third column as a sum of two columns:

\[ D = \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| \]

Let \( D_1 = \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| \) and \( D_2 = \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| \).

Thus, we have:

\[ D = D_1 + D_2 = 0 \]

### Step 3: Analyze \( D_1 \)
The determinant \( D_1 \) can be simplified by noticing that the first two columns are polynomial expressions in \( a, b, c \):

\[ D_1 = \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| \quad \text{(after interchanging columns)} \]

This determinant represents the Vandermonde determinant, which is non-zero if \( a, b, c \) are distinct.

### Step 4: Analyze \( D_2 \)
The determinant \( D_2 \) can be expressed as:

\[ D_2 = \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| \quad \text{(after interchanging columns)} \]

This determinant is also a Vandermonde determinant and is non-zero if \( a, b, c \) are distinct.

### Step 5: Set up the equation
Since \( D_1 + D_2 = 0 \) and both \( D_1 \) and \( D_2 \) are non-zero, we can conclude that:

\[ D_1 = -D_2 \]

### Step 6: Non-coplanarity condition
Given that the vectors are non-coplanar, the scalar triple product must be non-zero. Therefore, the condition \( 1 + abc = 0 \) must hold, leading to:

\[ abc = -1 \]

### Conclusion
Thus, the value of \( abc \) is:

\[ \boxed{-1} \]