= Solution
Because the sphere passes through the origin and has centre $P$, its radius is $|P|$. A point $X$ lies on it exactly when
$$
|X-P|^2=|P|^2,
$$
or equivalently
$$
2P\cdot X=|X|^2.
$$
With $P=\alpha A+\beta B+\gamma C$, the three required equations are the <Gram matrix> system
$$
\begin{pmatrix}
A\cdot A&A\cdot B&A\cdot C\\
B\cdot A&B\cdot B&B\cdot C\\
C\cdot A&C\cdot B&C\cdot C
\end{pmatrix}
\begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}
=\frac12
\begin{pmatrix}|A|^2\\|B|^2\\|C|^2\end{pmatrix}.
$$
For the specified vectors this becomes
$$
\alpha+\beta=\frac12,\qquad
\alpha+2\beta+\gamma=1,\qquad
\beta+5\gamma=\frac52.
$$
Thus
$$
\alpha=\frac12,\qquad\beta=0,\qquad\gamma=\frac12,
$$
and the centre is
$$
\boxed{P=\frac12A+\frac12C
=\left(\frac12,\frac12,1\right)}.
$$
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