If the equation
$$
x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0
$$

has exactly one real solution, then the value of $p$ is $\qquad$
Hint: Use factorization to transform the original equation into $x^{2}+3 x-p=0$ or $x^{2}+3 x-2 p=0$.
From the conditions, we have
$$
\left\{\begin{array} { l } 
{ \Delta _ { 1 } = 9 + 4 p = 0 , } \\
{ \Delta _ { 2 } = 9 + 8 p < 0 }
\end{array} \text { or } \left\{\begin{array}{l}
\Delta_{1}=9+4 p<0, \\
\Delta_{2}=9+8 p=0 .
\end{array}\right.\right.
$$

Answer: $-\frac{9}{4}$.