$$\begin{align*} 3p(x) &= (3x)^2 + 2c(3x) - (3c(a+b) + 3(a^2 + ab + b^2)) \\ &= (3x + c)^2 - (c^2 + 3c(a+b) + 3(a^2 + ab + b^2)) \\ &= (3x + c)^2 - \left( \left(c + \frac{3}{2}(a+b)\right)^2 - \frac{9}{4}(a+b)^2 + 3(a^2 + ab + b^2) \right) \\ &= (3x + c)^2 - \left( c + \frac{3}{2}(a+b) \right)^2 - \frac{12(a^2 + ab + b^2) - 9(a^2 + 2ab + b^2)}{4} \\ &= (3x + c)^2 - \left( c + \frac{3}{2}(a+b) \right)^2 - \frac{3(a^2 - 2ab + b^2)}{4} \\ &\geq - \left( \left(c + \frac{3}{2}(a+b)\right)^2 + \frac{3(a-b)^2}{4} \right), \end{align*}$$
for all real $x$, with equality iff $x = -\frac{c}{3}$. Thus
$$\frac{(2c + 3(a+b))^2 + 3(a-b)^2}{12}$$
is the minimum value of $p$. Since $a \neq b$, this is a negative real number.

Next
$$\begin{aligned} p(a) &= 3a^2 + 2c a - c(a+b) - (a^2 + ab + b^2) \\ &= 2a^2 - ab - b^2 + c(a-b) \\ &= (2a + b)(a-b) + c(a-b) \\ &= (a-b)(2a + b + c). \end{aligned}$$
And interchanging $a$ and $b$, $p(b) = (b-a)(2b + a + c)$. Hence
$$p(a) + p(b) = (a-b)(2a + b + c - (2b + a + c)) = (a-b)^2 > 0$$