Let A = [ cases 2 & 3 a & 0 cases ], a ∈ R be written as P + Q...

A = \left[\begin{cases}2 & 3 \\ a & 0\end{cases}\right], a \in R

Then,

{A}^{T} = \left[\begin{cases}2 & a \\ 3 & 0\end{cases}\right]

Now,

$P = \left(\dfrac{A + {A}^{T}}{2}\right)$

⇒ P = \dfrac{1}{2}\left\{\left[\begin{cases}2 & 3 \\ a & 0\end{cases}\right] + \left[\begin{cases}2 & a \\ 3 & 0\end{cases}\right]\right\}

⇒ P = \left[\begin{cases}2 & \dfrac{3 + a}{2} \\ \dfrac{a + 3}{2} & 0\end{cases}\right]
And,

$Q = \dfrac{A - {A}^{T}}{2}$

⇒ Q = \dfrac{1}{2}\left\{\left[\begin{cases}2 & 3 \\ a & 0\end{cases}\right] - \left[\begin{cases}2 & a \\ 3 & 0\end{cases}\right]\right\}

⇒ Q = \left[\begin{cases}0 & \dfrac{3 - a}{2} \\ \dfrac{a - 3}{2} & 0\end{cases}\right]

As,

$det ( Q ) = 9$

$⇒ {\left(a - 3\right)}^{2} = 36$

$⇒ a = 3 \pm 6$

$∴ a = 9, - 3$

Now,

$det \left(P\right) = - {\left(\dfrac{3 + a}{2}\right)}^{2}$

For $a = - 3, det \left(P\right) = 0$

For $a = 9$, $det \left(P\right) = - {\left(\dfrac{3 + 9}{2}\right)}^{2} = - 36$
$∴$ Modulus of the sum of all possible values of $det\left(P\right)$

$= \left|- 36\right| + \left|0\right| = 36$