If $y=-4$ is a root of $\begin{vmatrix}y&2&3\\1&y&1\\3&2&y\end{vmatrix}$, then the product of the other two roots is

3

The correct answer is Option (4) → 3

Given determinant: $D = \begin{vmatrix} y & 2 & 3 \\ 1 & y & 1 \\ 3 & 2 & y \end{vmatrix}$

Expand along first row:

$D = y \begin{vmatrix} y & 1 \\ 2 & y \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 3 & y \end{vmatrix} + 3 \begin{vmatrix} 1 & y \\ 3 & 2 \end{vmatrix}$

$= y(y^2 - 2) - 2(y - 3) + 3(2 - 3y)$

$= y^3 - 2y - 2y + 6 + 6 - 9y = y^3 - 13y + 12$

So characteristic equation: $y^3 - 13y + 12 = 0$

Given one root $y = -4$

Factor: $(y + 4)(y^2 - 4y + 3) = y^3 - 13y + 12$ (check: $y^2 -4y +3$ → roots of remaining two)

Product of other two roots = $3$ (constant term of quadratic)