Positive definiteness depending on a parameter

Matrix.
$$A=\begin{pmatrix}1&a&a\\a&1&a\\a&a&1\end{pmatrix}.$$

Leading minors.
- $\Delta_1=1>0$.
- $\Delta_2=\begin{vmatrix}1&a\\a&1\end{vmatrix}=1-a^2>0\iff|a|<1$.
- $\Delta_3=\det A$. Compute: add all rows to the first, factor $(1+2a)$, then $\Delta_3=(1+2a)(1-a)^2$. So $\Delta_3>0\iff a>-1/2$ and $a\neq1$.

Combining. $|a|<1$ and $a>-1/2$ gives $-1/2<a<1$. At $a=1$, $\Delta_3=0$ (rank $1$), not positive definite. At $a=-1/2$, $\Delta_3=0$ (degenerate), not positive definite.

Conclusion. $Q$ is positive definite iff $-1/2<a<1$.