MHTCET 5 May 2025 Shift 2 Paper

Concept:Non-coplanar vectors have a non-zero scalar triple product; the given determinant condition then forces $abc + 1 = 0$.Explanation:Since $\overline{\mathrm{p}}, \overline{\mathrm{q}}, \overline{\mathrm{r}}$ are non-coplanar, their scalar triple product is non-zero:$$D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \neq 0.$$The given condition is:$$\begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{vmatrix} = 0.$$Split the third column using linearity of determinants:$$\begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} + \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = 0.$$Take $a, b, c$ common from $R_1, R_2, R_3$ respectively in the first determinant:$$abc\,D + \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = 0.$$In the second determinant, apply $C_2 \leftrightarrow C_3$ and then $C_1 \leftrightarrow C_2$. Since each interchange changes the sign, two interchanges give no net sign change and it becomes $D$:$$abc\,D + D = 0.$$$$\Rightarrow (abc + 1)D = 0.$$Since $D \neq 0$, we get:$$abc + 1 = 0 \Rightarrow abc = -1.$$Answer:$$abc = -1$$Hence, the correct option is B.