7. Let $a=\sqrt{3 x+1}+\sqrt{3 y+1}+\sqrt{3 z+1}$, where $x+y+z=1, x, y, z \geqslant 0$. Then $[a]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
7.4.

Notice,
$$
\begin{aligned}
a^{2}= & (3 x+1)+(3 y+1)+(3 z+1)+ \\
& 2 \sqrt{(3 x+1)(3 y+1)}+ \\
& 2 \sqrt{(3 y+1)(3 z+1)}+ \\
& 2 \sqrt{(3 z+1)(3 x+1)} \\
\leqslant & 3((3 x+1)+(3 y+1)+(3 z+1))=18 .
\end{aligned}
$$

Then $a \leqslant \sqrt{18}<5$.
Also, $0 \leqslant x, y, z \leqslant 1$, so,
$$
x \geqslant x^{2}, y \geqslant y^{2}, z \geqslant z^{2} \text {. }
$$
$$
\begin{array}{l}
\text { Hence } a \geqslant \sqrt{x^{2}+2 x+1}+\sqrt{y^{2}+2 y+1}+\sqrt{z^{2}+2 z+1} \\
=(x+1)+(y+1)+(z+1)=4 .
\end{array}
$$

Therefore, $4 \leqslant a<5 \Rightarrow[a]=4$.