No, the answer is not $3\sqrt{3}$

I will solve this by reducing the number of variables in the problem: By Lagrange multipliers, an extreme value must satisfy the equation $(2 - bc, 2 - ac, 2 - ba) = \lambda (2a,2b,2c)$ .

Case 1: At least one of the variables equals 0. Then let's say $c = 0$ . If that is the case, we have to solve the problem $a^2 + b^2 = 9 \\ \text{Max: }2(a+b)$

Case 2: All variables are non-zero. Is that is the case, we may divide by the variables and by the previous equation, we find that $\frac{2 - bc}{a} = \frac{2 - ac}{b} = \frac{2 - ba}{c}$ .

Then, for example, $\frac{2 - ac}{b} = \frac{2-bc}{a} \implies 2a - a^2c = 2b - b^2c \implies 2(a-b) = c(a^2 - b^2)$ . Similarly, by symmetry we can pick other pairs of equations and prove that $2(b-c) = a(b^2 - c^2), 2(a-c) = b(a^2 - c^2)$ . Let's now assume that no variables are equal. Then we may divide by terms of the form $a - b$ establishing that $2 = c(a+b) = b(a+c) = a(b+c)$ . Then $c(a+b) = b(a+c) \implies ca + cb = ba + bc \implies ca = ba \implies c = b$ . This is a contradiction, because we assumed no variables were equal. This means that in this case, it must be the case that at least two variables are equal.

Let's assume that $c = a$ . Then we have to solve $2a^2 + b^2 = 9 \\ \text{Max: } 2(2a + b) - a^2b$ .

Thus we have reduced the problem into solving two problems with just two variables. Two variable inequalities are the simplest so I will skip a complete proof (though I may add one if someone wants me to) and will just present the results. The answer to the problem in Case 1 is $\frac{12}{\sqrt{2}} \approx 8$ and the answer for the problem in Case 2 is $10$ . Comparing the two numbers, it is clear the solution is $10$ .

This is not full solution ...

let $F(a,b,c)= 2(a+b+c)-abc$ and w.l.o.g. $a ≥b ≥c$

There are 3 cases.

1) c=0 : easy

2) c>0 : can't get max because $F(a,b,c) - F(a,b,-c) = 4c - 2abc = 2c( 2-ab )$ : if $ab>2 : F(a,b,c) < F(a,b,-c)$

3) c<0

$F(a,b,c) - F( \frac{a+b} {2} , \frac{a+b} {2} , c ) = c ( \frac {a-b}{2} )^2 < 0$

So, let's do a=b , and then, by differential, we can get max when a=b=2, c=-1

$a=2,b=2,c=-1$