If $y = -4$ is a root of $\begin{vmatrix} y & 2 & 3 \\ 1 & y & 1 \\ 3 & 2 & y \end{vmatrix} = 0$, then the product of the other two roots is

The determinant equation is:

$\begin{vmatrix} y & 2 & 3 \\ 1 & y & 1 \\ 3 & 2 & y \end{vmatrix} = 0$

Expanding along the first row:

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

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

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

$= y^3 - 13y + 12$

$= 0$

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For a cubic equation $y^3 + ay^2 + by + c = 0$ with roots $\alpha, \beta, \gamma$, Vieta's formula states:

$\alpha \times \beta \times \gamma = -c$

The equation is $y^3 + 0y^2 - 13y + 12 = 0$, so $c = 12$.

Product of all three roots $= -12$

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If the three roots are $-4, \alpha, \beta$, then:

$(-4) \times \alpha \times \beta = -12$

$\alpha \times \beta = \dfrac{-12}{-4}$

$\alpha \times \beta = 3$

Therefore, the product of the other two roots is $3$.