If $\begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$, then product of all values of $a$ is:

The determinant is given as:

$\begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$

Expanding along Row 1:

$\text{Det} = 1 \cdot M_{11} - (-2) \cdot M_{12} + 5 \cdot M_{13}$

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For $M_{11}$, remove row 1 and column 1:

$M_{11} = \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix}$

$= a(2a) - (-1)(4)$

$= 2a^2 + 4$

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For $M_{12}$, remove row 1 and column 2:

$M_{12} = \begin{vmatrix} 2 & -1 \\ 0 & 2a \end{vmatrix}$

$= 2(2a) - (-1)(0)$

$= 4a$

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For $M_{13}$, remove row 1 and column 3:

$M_{13} = \begin{vmatrix} 2 & a \\ 0 & 4 \end{vmatrix}$

$= 2(4) - a(0)$

$= 8$

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Substituting into the determinant formula:

$\text{Det} = 1(2a^2 + 4) + 2(4a) + 5(8)$

$= 2a^2 + 4 + 8a + 40$

$= 2a^2 + 8a + 44$

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Setting the determinant equal to 86:

$2a^2 + 8a + 44 = 86$

$2a^2 + 8a - 42 = 0$

$a^2 + 4a - 21 = 0$

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Factoring the quadratic:

$(a + 7)(a - 3) = 0$

Therefore:

$a = -7$ or $a = 3$

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The product of all values of $a$:

$(-7) \times 3 = -21$