3. Given the seven-variable polynomial
$$
\begin{array}{l}
Q\left(x_{1}, x_{2}, \cdots, x_{7}\right) \\
=\left(x_{1}+x_{2}+\cdots+x_{7}\right)^{2}+2\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{7}^{2}\right)
\end{array}
$$

it can be expressed as the sum of seven squares of polynomials with non-negative integer coefficients, i.e.,
$$
\begin{array}{l} 
Q\left(x_{1}, x_{2}, \cdots, x_{7}\right) \\
= P_{1}^{2}\left(x_{1}, x_{2}, \cdots, x_{7}\right)+ \\
P_{2}^{2}\left(x_{1}, x_{2}, \cdots, x_{7}\right)+\cdots+ \\
P_{7}^{2}\left(x_{1}, x_{2}, \cdots, x_{7}\right),
\end{array}
$$

where $P_{k}\left(x_{1}, x_{2}, \cdots, x_{7}\right)$ are polynomials with non-negative integer coefficients. Find all possible values of $P_{1}(1,1, \cdots, 1)$.
3. When $x_{1}=x_{2}=\cdots=x_{7}=0$,
$$
\begin{aligned}
0= & Q(0,0, \cdots, 0) \\
= & P_{1}^{2}(0,0, \cdots, 0)+P_{2}^{2}(0,0, \cdots, 0)+\cdots+ \\
& P_{7}^{2}(0,0, \cdots, 0) .
\end{aligned}
$$

Thus, the constant terms of $P_{k} (k=1,2, \cdots, 7)$ are all 0.
Notice that, if the degree of some $P_{k}$ is greater than 1, then
$P_{k}$ contains a term $a x_{1}^{\alpha_{1}} x_{2}^{\alpha_{2}} \cdots x_{7}^{\alpha_{7}}\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{7}>1\right)$.

Therefore, the term $x_{1}^{2 \alpha_{1}} x_{2}^{2 \alpha_{2}} \cdots x_{7}^{2 \alpha_{7}}$ in $P_{1}^{2}+P_{2}^{2}+\cdots+P_{7}^{2}$ has a degree greater than 2, which contradicts the fact that the degree of $Q(x)$ is 2.
Thus, the degree of each $P_{k}$ does not exceed 1.
Since the degree of $P_{k}$ is 1 and its constant term is 0, we have
$$
\begin{array}{l}
P_{k}=a_{k 1} x_{1}+a_{k 2} x_{2}+\cdots+a_{k 7} x_{7} . \\
\text { Also, } Q(1,0,0, \cdots, 0)=3, \text { then } \\
P_{1}^{2}(1,0,0, \cdots, 0)+P_{2}^{2}(1,0,0, \cdots, 0)+\cdots+ \\
P_{7}^{2}(1,0,0, \cdots, 0) \\
=a_{11}^{2}+a_{21}^{2}+\cdots+a_{71}^{2}=3 .
\end{array}
$$

Since $a_{11}, a_{21}, \cdots, a_{71}$ are all non-negative integers, exactly three of them equal 1, and the rest equal 0.

Similarly, for each $k (k=1,2, \cdots, 7)$, the coefficients of $x_{k}$ are such that exactly three equal 1, and the other four equal 0.

Thus, the sum of all coefficients of $P_{1}, P_{2}, \cdots, P_{7}$ is $3 \times 7=21$, i.e., $\square$
$$
\sum_{i=1}^{7} P_{i}(1,1, \cdots, 1)=21 \text {. }
$$

Also, since $Q(1,1, \cdots, 1)=63$, i.e.,
$$
P_{1}(1,1, \cdots, 1), P_{2}(1,1, \cdots, 1), \cdots, P_{7}(1,1, \cdots, 1)
$$

the squares of these seven numbers sum to 63, and $\frac{21}{7}=\sqrt{\frac{63}{7}}$, so by the mean inequality,
$$
\begin{array}{l}
P_{1}(1,1, \cdots, 1)=P_{2}(1,1, \cdots, 1) \\
=\cdots=P_{7}(1,1, \cdots, 1)=3 .
\end{array}
$$