Note that if $a$ is not an eigenvalue of the matrix $A$, then the matrix $A-aI$ is nonsingular, that is, the null space of $A-aI$ is the zero space, and the dimension is $0$.
Thus to obtain a nonzero point, your favorite number should be an eigenvalue.

Let us find the eigenvalue. The characteristic polynomial $p(t)$ of $A$ is
\begin{align*}
p(t)&=\det(A-tI)\\
&\begin{vmatrix}
5-t & 2 & -1 \\
2 &2-t &2 \\
-1 & 2 & 5-t
\end{vmatrix}\\
&=(5-t)\begin{vmatrix}
2-t & 2\\
2& 5-t
\end{vmatrix}-2\begin{vmatrix}
2 & 2\\
-1& 5-t
\end{vmatrix}+(-1)\begin{vmatrix}
2 & 2-t\\
-1& 2
\end{vmatrix}\\
&\text{(By the first row cofactor expansion.)}\\
&=(5-t)(t^2-7t+6)-2(-2t+12)-(-t+6)\\
&=-t^3+12t^2-36t\\
&=-t(t-6)^2.
\end{align*}

Eigenvalues are roots of the characteristic polynomial.
Thus eigenvalues of $A$ are $0$ and $6$, with algebraic multiplicities $1$ and $2$, respectively.

Since the matrix $A$ is a real symmetric matrix, it is diagonalizable. Thus the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.

Therefore the dimension of the null space of $A-6I$, which is the geometric multiplicity of the eigenvalue $6$ is equal to $2$, and the dimension of the null space of $A$ is $1$.

Thus, if you take $a=6$, you get $2\times 5=10$ points. If your favorite number is $a=0$, then your score is $1\times 5=5$ points. The other choice of $a$ will be zero points.

Comment.

If you didn’t notice that the given matrix is a real symmetric matrix, then you need to find eigenvectors corresponding to eigenvalues $6$ and to show that the eigenspace has dimension $2$.

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