The vectors $\overline{\mathrm{p}}=\hat{\mathrm{i}}+a \hat{\mathrm{j}}+a^2 \hat{\mathrm{k}},…

The determinant $D$ is given by:

$D = \left|\begin{array}{lll} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{array}\right| = 0$

Expressing the third column as a sum of two columns yields:

$D = \left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right| + \left|\begin{array}{lll} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array}\right| = 0$

The first determinant is a Vandermonde determinant:
$\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = (a-b)(b-c)(c-a)$

Factoring $a$, $b$, and $c$ from the rows of the second determinant gives:
$abc \cdot \left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = abc(a-b)(b-c)(c-a)$

Substituting into the original equation provides:
$(a-b)(b-c)(c-a) + abc(a-b)(b-c)(c-a) = 0$

Factoring yields:
$(a-b)(b-c)(c-a)(1 + abc) = 0$

The vectors $\overline{\mathrm{p}}$, $\overline{\mathrm{q}}$, and $\overline{\mathrm{r}}$ are non-coplanar, so their scalar triple product $[\overline{\mathrm{p}}\ \overline{\mathrm{q}}\ \overline{\mathrm{r}}] = (a-b)(b-c)(c-a) \neq 0$.
Thus, $1 + abc = 0$, giving $abc = -1$.

The result is $\boxed{-1}$.