7. In the Cartesian coordinate system $x O y$, it is known that the graph of the parabola $y=a x^{2}-3 x+3(a \neq 0)$ is symmetric to the graph of the parabola $y^{2}=2 p x(p>0)$ with respect to the line $y=x+m$. Then the product of the real numbers $a, p, m$ is $\qquad$ .
Let $M\left(x_{0}, y_{0}\right)$ be a point on the parabola $y^{2}=2 p x$, and its symmetric point with respect to the line $y=x+m$ be
$$
\begin{array}{l}
M^{\prime}\left(y_{0}-m, x_{0}+m\right) . \text { Then we have }\left\{\begin{array}{l}
y_{0}^{2}=2 p x_{0}, \\
x_{0}+m=a\left(y_{0}-m\right)^{2}-3\left(y_{0}-m\right)+3
\end{array}\right. \\
\Rightarrow y_{0}^{2}-\frac{2 a m+3}{a} y_{0}+\frac{a m^{2}+2 m+3}{a}=\frac{x_{0}}{a} \\
\Rightarrow\left\{\begin{array} { l } 
{ 2 p = \frac { 1 } { a } , } \\
{ 2 a m + 3 = 0 , } \\
{ a m ^ { 2 } + 2 m + 3 = 0 }
\end{array} \quad \Rightarrow \left\{\begin{array}{l}
a=\frac{1}{4}, \\
m=-6, \\
p=2
\end{array}\right.\right.
\end{array}
$$