Let $f(x)=\{\begin{array}{lll}{2}^{\frac{\sin <!--\; \; -->(x-<!--\; -\; -->3)}{x-<!--\; -\; -->[x]}}& ;& x<3\\ \frac{a|-<!--\; -\; -->x^2-<!--\; -\; -->12+7|}{b(x^2+12x-<!--\; -\; -->7)}& ;& x>3\\ b& ;& x=3\end{array}$ is continuous at x = 3, then (a, b) is

The correct option is None of these.

Underlying Concept:
For a function $f\left(x\right)$ to be continuous at a point $x=c$, the Left-Hand Limit (LHL), the Right-Hand Limit (RHL), and the function value at that point must all exist and be equal:
$$\underset{x\to c^{-}}{lim}f(x)=\underset{x\to c^{+}}{lim}f(x)=f(c)$$

Here, we are given that the function:
$$f(x)=\{\begin{array}{lll}{2}^{\frac{\sin (x-3)}{x-[x]}}& ;& x<3\\ \frac{a|-x^2-12x+7|}{b(x^2+12x-<!--\; -\; -->7)}& ;& x>3\\ b& ;& x=3\end{array}$$
is continuous at $x=3$.

Step 1: Find the Left-Hand Limit (LHL) at $x=3$
For $x<3$, let $x=3-h$ where $h\to 0$ $(h>0)$.
The expression in the exponent is:
$$\frac{\sin (x-3)}{x-[x]}$$
Here, $[x]$ represents the greatest integer function (floor function).
As $x\to 3^{-}$, $x$ is slightly less than 3 (e.g., $2.999$), so:
$$[x]=2$$
Thus, the denominator becomes:
$$x-[x]=x-2$$
Substituting $x=3$ into the limit:
$$\underset{x\to 3^{-}}{lim}\frac{\sin (x-3)}{x-2}=\frac{\sin (3-3)}{3-2}=\frac{\sin (0)}{1}=0$$
Therefore, the Left-Hand Limit is:
$$\text{LHL}=\underset{x\to 3^{-}}{lim}f(x)=2^0=1$$

Step 2: Relate LHL to the value of the function at $x=3$
Since the function is continuous at $x=3$:
$$\text{LHL}=f(3)\Rightarrow 1=b$$
Thus, we obtain:
$$b=1$$

Step 3: Find the Right-Hand Limit (RHL) at $x=3$
For $x>3$, let $x=3+h$ where $h\to 0$ $(h>0)$.
The expression for RHL is:
$$\text{RHL}=\underset{x\to 3^{+}}{lim}\frac{a|-x^2-12x+7|}{b(x^2+12x-7)}$$
Let us evaluate the sign of the term inside the absolute value, $g\left(x\right)=-x^2-12x+7$, as $x\to 3^{+}$:
$$g(3)=-3^2-12(3)+7=-9-36+7=-38$$
Since $-38<0$, the expression inside the absolute value is strictly negative in a neighborhood around $x=3$.
Using the definition of absolute value $|y|=-y$ when $y<0$:
$$|-x^2-12x+7|=-(-x^2-12x+7)=x^2+12x-7$$
Now, substitute this simplification back into the limit:
$$\text{RHL}=\underset{x\to 3^{+}}{lim}\frac{a}{b}·\frac{x^2+12x-7}{x^2+12x-7}=\frac{a}{b}(1)=\frac{a}{b}$$

Step 4: Solve for the parameters
For continuity, the Right-Hand Limit must equal the Left-Hand Limit:
$$\text{RHL}=\text{LHL}\Rightarrow \frac{a}{b}=1\Rightarrow a=b$$
Since we already established $b=1$, it follows that:
$$a=1$$
Thus, the ordered pair $(a,b)$ is:
$$(a,b)=(1,1)$$

Comparing $(1,1)$ with the given choices:
1. $(2,3)$
2. $(1,2)$
3. $(2^{\sin 2},3)$
Since $(1,1)$ is not present among these options, the correct choice is None of these.