Answer
9, 9, 9
Work Step by Step
Let $a^{2}+b^{2}+c^{2}=f(a, b, c)$ be our function where we know that $27=a+b+c$. Taking into account that
\[
\begin{aligned}
g(a, b, 27-a-b)=g(a, b)=a^{2}=g(a, b, c)=a^{2}+b^{2}+(27-a-b)^{2} \\
g(a, b)=2 a^{2}+2 b^{2}+2 a b-54(a+b)+27^{2}
\end{aligned}
\]
We can use the first derivative test:
\[
\begin{aligned}
g_{a}=0 & \Rightarrow 4 a+2 b-54=0 \text { and } g_{b}=0 \Rightarrow 4 b+2 a-54=0 \\
& \Rightarrow\left\{\begin{array}{l}
4 a+2 b-54=0 \\
-8 b-4 a+108=0
\end{array}\right.\\
\Rightarrow-6 b+54=0 \Rightarrow b=54 / 6=9
\end{aligned}
\]
Now we can use the second derivative test for the critical point $P(9,9,9)$
\[
\begin{array}{r}
g_{\text {oa }}=g_{t b}=4 \text { and } g_{\text {ab }}=2 \\
\Rightarrow D=g_{\text {aa }} g_{\text {bb }}-g_{\text {ab }}^{2}=4(4)-4>0 \text { and } g_{\text {an }}=4>0
\end{array}
\]
Since $D, g_{0 a}>0, g(a, b, c)$ has a absolute minimum at $P(9,9,9)$.
