3. Let real numbers $a, b, c \geqslant 1$, and satisfy $a b c+2 a^{2}+2 b^{2}+2 c^{2}+c a-c b-4 a+4 b-c=28$, find the maximum value of $a+b+c$.

The above text is translated into English, please retain the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
a b c+2 a^{2}+2 b^{2}+2 c^{2}+c a-c b-4 a+4 b-c=2 a^{2}+2 b^{2}+2 c^{2}-4 a+4 b+c(a b+a-b-1) \\
=2(a-1)^{2}+2(b+1)^{2}+2 c^{2}+c(a-1)(b+1)-4=28 \\
\Rightarrow 2(a-1)^{2}+2(b+1)^{2}+2 c^{2}+c(a-1)(b+1)=32,
\end{array}
$$

Let $x=\frac{a-1}{2}, y=\frac{b+1}{2}, z=\frac{c}{2}$, then the above equation can be equivalently transformed into $x^{2}+y^{2}+z^{2}+x y z=4$.
Assume $x \geqslant y \geqslant z>0$, then $4=x^{2}+y^{2}+z^{2}+x y z \leqslant 3 x^{2}+x^{3} \Rightarrow x^{3}+3 x^{2}-4 \geqslant 0$ $\Rightarrow(x-1)(x+2)^{2} \geqslant 0 \Rightarrow x \geqslant 1$, similarly, we can get $z \leqslant 1$.
Thus, among $x, y, z$, there must be two numbers that are both not less than 1 or both not greater than 1, assume they are $y, z$, then $(y-1)(z-1) \geqslant 0 \Rightarrow y z-y-z+1 \geqslant 0 \Rightarrow y z \geqslant y+z-1$.
Therefore, $4-x^{2}=y^{2}+z^{2}+x y z \geqslant 2 y z+x y z=y z(2+x) \Rightarrow 2-x \geqslant y z \geqslant y+z-1$ $\Rightarrow x+y+z \leqslant 3 \Rightarrow \frac{a-1}{2}+\frac{b+1}{2}+\frac{c}{2} \leqslant 3 \Rightarrow a+b+c \leqslant 6$,
Equality holds when $x=y=z=1 \Rightarrow a=3, b=1, c=2$.