Question:

<p><strong>Step 1: Decompose the given determinant</strong></p><p>We can write the third column as a sum:</p><p>$$\begin{vmatrix} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{vmatrix} = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix}$$</p><p><strong>Step 2: Evaluate the Vandermonde determinant</strong></p><p>The first determinant (with columns 1, a, a²) is the Vandermonde determinant:</p><p>$$\begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$$</p><p>Since vectors (1, a, a²), (1, b, b²), (1, c, c²) are non-coplanar, this determinant is non-zero, meaning $a$, $b$, $c$ are all distinct.</p><p><strong>Step 3: Factor the second determinant</strong></p><p>For the second determinant, factor out $a$, $b$, $c$ from respective rows:</p><p>$$\begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = abc \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = abc(a-b)(b-c)(c-a)$$</p><p><strong>Step 4: Apply the given condition</strong></p><p>Given that the original determinant equals 0:</p><p>$$(a-b)(b-c)(c-a) + abc(a-b)(b-c)(c-a) = 0$$</p><p>Factor out $(a-b)(b-c)(c-a)$:</p><p>$$(a-b)(b-c)(c-a)[1 + abc] = 0$$</p><p><strong>Step 5: Use non-coplanarity</strong></p><p>Since the vectors are non-coplanar, $(a-b)(b-c)(c-a) \neq 0$</p><p>Therefore: $1 + abc = 0$</p><p>$$abc = -1$$</p><p><strong>∴ Answer: B</strong></p>