1. We are given $A=\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}$ and $AA^T=9I.$

2. Since $AA^T=9I$, the rows of $A$ must be pairwise orthogonal and each row must have magnitude $3$.

Let the rows be: $R_1=(1,2,2),\quad R_2=(2,1,-2),\quad R_3=(a,2,b).$

Then:

• $R_i\cdot R_i=9$ for each $i$
• $R_i\cdot R_j=0$ for $i\ne j$

3. First check the first two rows: $R_1\cdot R_1=1^2+2^2+2^2=1+4+4=9,$ $R_2\cdot R_2=2^2+1^2+(-2)^2=4+1+4=9,$ $R_1\cdot R_2=1\cdot2+2\cdot1+2\cdot(-2)=2+2-4=0.$

So the first two rows already satisfy the condition.

4. Now impose orthogonality of $R_3=(a,2,b)$ with $R_1$ and $R_2$.

From $R_1\cdot R_3=0$: $1\cdot a+2\cdot2+2\cdot b=0$ $a+4+2b=0$ $a+2b=-4 \qquad (1)$

From $R_2\cdot R_3=0$: $2\cdot a+1\cdot2+(-2)\cdot b=0$ $2a+2-2b=0$ $a-b=-1 \qquad (2)$

5. Solve equations (1) and (2).

From (2): $a=b-1.$

Substitute into (1): $(b-1)+2b=-4$ $3b-1=-4$ $3b=-3$ $b=-1.$

Then $a=b-1=-1-1=-2.$

So $(a,b)=(-2,-1).$

6. Verify length of third row: $R_3\cdot R_3=(-2)^2+2^2+(-1)^2=4+4+1=9.$

Hence all conditions are satisfied.

Therefore the correct option is $\boxed{(-2,-1)}$ which is Option B.

7. Comparison with stored correct answer: Stored correct answer = B. Our derived answer = B. So they agree.