Let $\alpha, \beta, \gamma$ be the roots of $p(x)=0$.
By Vieta's formulas, we have
$$\left\{\begin{array}{l} \alpha+\beta+\gamma=-a, \\ \alpha \beta+\beta \gamma+\gamma \alpha=b, \\ \alpha \beta \gamma=-c . \end{array}\right.$$

Then $a^{2}-2 b=\alpha^{2}+\beta^{2}+\gamma^{2}$.
The original inequality
$$\begin{aligned} \Leftrightarrow & 6 a\left(a^{2}-2 b\right)+10\left(a^{2}-2 b\right)^{\frac{3}{2}} \geqslant 27 c \\ \Leftrightarrow & 6(\alpha+\beta+\gamma)\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)- \\ & 10\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)^{\frac{3}{2}} \leqslant 27 \alpha \beta \gamma . \end{aligned}$$

When $\alpha^{2}+\beta^{2}+\gamma^{2}=0$, $\alpha=\beta=\gamma=0$, i.e., $a=b=c=0$.

In this case, inequality (1) clearly holds, and equality holds.
Assume $\alpha^{2}+\beta^{2}+\gamma^{2}>0$.
Since inequality (1) is a homogeneous cubic expression, to eliminate the 27 on the right side of (1), we can assume $\alpha^{2}+\beta^{2}+\gamma^{2}=9$. At this point, inequality (1) simplifies to
$$2(\alpha+\beta+\gamma)-\alpha \beta \gamma \leqslant 10 .$$

Assume $|\alpha| \leqslant|\beta| \leqslant|\gamma|$. Then
$$\gamma^{2} \geqslant 3 \text {. }$$

Thus, $2 \alpha \beta \leqslant \alpha^{2}+\beta^{2}=9-\gamma^{2} \leqslant 6$.
Consider the square of the left side of (2)
$$\begin{array}{l} (2(\alpha+\beta+\gamma)-\alpha \beta \gamma)^{2} \\ =(2(\alpha+\beta)+(2-\alpha \beta) \gamma)^{2} \\ \leqslant\left(4+(2-\alpha \beta)^{2}\right)\left((\alpha+\beta)^{2}+\gamma^{2}\right) \\ =\left(8-4 \alpha \beta+(\alpha \beta)^{2}\right)(9+2 \alpha \beta) \\ =2(\alpha \beta)^{3}+(\alpha \beta)^{2}-20 \alpha \beta+72 \\ =(\alpha \beta+2)^{2}(2 \alpha \beta-7)+100 \\ \leqslant 100 \\ \Rightarrow 2(\alpha+\beta+\gamma)-\alpha \beta \gamma \leqslant 10 . \end{array}$$

Therefore, the original inequality holds.
If equality holds in (2), then equality must hold in (4), which implies $\alpha \beta=-2$.
Substituting into (2) gives
$$\begin{array}{l} \alpha+\beta+2 \gamma=5 \\ \Rightarrow(\alpha+\beta)^{2}=(5-2 \gamma)^{2} . \\ \text { By }(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta \\ =\left(9-\gamma^{2}\right)-4=5-\gamma^{2} \\ \Rightarrow 5-\gamma^{2}=(5-2 \gamma)^{2}=25-20 \gamma+4 \gamma^{2} \\ \Leftrightarrow \gamma^{2}-4 \gamma+4=0 \\ \Leftrightarrow \gamma=2 . \end{array}$$

Substituting into (5) gives $\alpha+\beta=1$.
Then $\alpha, \beta$ are the roots of the quadratic equation $x^{2}-x-2=0$.
Thus, $\alpha=-1, \beta=2$.
Since (1) is a homogeneous cubic expression, we can assume
$$p(x)=x^{3}+a x^{2}+b x+c$$

has three real roots $-\lambda, 2 \lambda, 2 \lambda(\lambda>0)$.
By Vieta's formulas, we have
$$\begin{array}{l} a=-3 \lambda, b=0, \\ c=4 \lambda^{3}=\frac{4}{27}(3 \lambda)^{3}=-\frac{4}{27} a^{3} . \end{array}$$

In summary, the necessary and sufficient condition for equality in the original inequality is $b=0, c=-\frac{4}{27} a^{3}(a \leqslant 0)$.