= Solution
Take the usual symmetric meaning of positive definite $Q$. To include all data allowed by the source, first check feasibility. Put
$$
B=AQ^{-1}A^T,\qquad x_0=Q^{-1}A^TB^{-1}b,\qquad
\rho^2=b^TB^{-1}b.
$$
Full row rank makes $B$ positive definite. Every $x$ satisfying $Ax=b$ is uniquely $x_0+z$ with $Az=0$, and $z^TQx_0=0$. Thus
$$
x^TQx=\rho^2+z^TQz.
$$
If $\rho^2>1$, there is no feasible point. If $\rho^2=1$, the only feasible point is $x_0$, and \b[$x_0$ is optimal for every $c$]. If $\rho^2<1$, the <Slater condition> holds, and the following <KKT conditions> are necessary and sufficient:
$$
\boxed{Ax=b,\quad x^TQx\leq1,\quad \gamma\geq0,\quad
c-A^Ty+2\gamma Qx=0,\quad\gamma(x^TQx-1)=0.}
$$
Necessity follows from the <constraint qualification> proved in (ii). Sufficiency follows from <convexity>: the stationary <optimization Lagrangian> $c^Tx+y^T(b-Ax)+\gamma(x^TQx-1)$ is minimized at $x$, while its feasible values never exceed the objective, and <complementary slackness> makes its value at $x$ equal the objective.
The singleton case cannot be silently covered by ordinary KKT necessity. For example, let $Q=I$, $A=(1\ 0)$, $b=1$, $c=(0,1)^T$. The sole feasible point $(1,0)$ is optimal, but stationarity would require $(0,1)$ to lie in the span of $(1,0)$, which is impossible. The feasibility split above supplies genuinely necessary and sufficient conditions without assuming an unprinted regularity hypothesis.
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