5. Given that $a, b, c$ are the lengths of the three sides of a triangle, and real numbers $p, q$ satisfy $p+q=1$. Then the result of $p a^{2}+q b^{2}-p q c^{2}$ is $(\quad)$.
(A) Positive
(B) Zero
(C) Negative
(D) Any of the above is possible
5. A.

Let $y=p a^{2}+q b^{2}-p q c^{2}$.
Substitute $q=1-p$ into the above equation:
$$
\begin{array}{l}
y=p a^{2}+(1-p) b^{2}-p(1-p) c^{2} \\
=c^{2} p^{2}+\left(a^{2}-b^{2}-c^{2}\right) p+b^{2} .
\end{array}
$$

Consider the above equation as a quadratic function of $p$, with the parabola opening upwards, then
$$
\begin{array}{l}
\Delta=\left(a^{2}-b^{2}-c^{2}\right)^{2}-4 b^{2} c^{2} \\
=\left[a^{2}-(b-c)^{2}\right]\left[a^{2}-(b+c)^{2}\right] . \\
=(a+b-c)(a-b+c)(a+b+c)(a-b-c) .
\end{array}
$$

Since $a, b, c$ are the sides of a triangle, we have
$$
\begin{array}{l}
a+b-c>0, a-b+c>0, \\
a+b+c>0, a-b-c<0 .
\end{array}
$$

Therefore, $\Delta<0$.