$$\begin{gathered} \mathrm{If}\underset{x\to 2}{lim}\frac{\sin (x^3-5x^2+ax+b)}{(\sqrt{x-1}-1).{\log }_e(x-1)}=m\mathrm{(exists}\mathrm{finitely) } \\ \mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}a+b+m \end{gathered}$$

The correct answer is 6.

Given the limit:
$\underset{x\to 2}{lim}\frac{\sin (x^3-5x^2+ax+b)}{(\sqrt{x-1}-1){\log }_e(x-1)}=m$
where $m$ exists and is finite.

Step 1: Analyze the behavior of the denominator as $x\to 2$
Let $t=x-2$. As $x\to 2$, we have $t\to 0$.
Rewriting the terms in the denominator in terms of $t$:
1) $\sqrt{x-1}-1=\sqrt{t+1}-1$
Using the Taylor expansion or rationalization, as $t\to 0$:
$\sqrt{t+1}-1\approx \frac{1}{2}t$
2) ${\log }_e(x-1)={\log }_e(t+1)$
As $t\to 0$:
${\log }_e(t+1)\approx t$
Thus, the denominator behaves like:
$(\sqrt{x-1}-1){\log }_e(x-1)\approx \frac{1}{2}{t}^2=\frac{1}{2}{(x-2)}^2$

Step 2: Determine conditions on the numerator for a finite limit
For the limit to exist and be finite, the numerator must vanish at least to the same order as the denominator at $x=2$.
Since the denominator has a zero of order 2 at $x=2$, the numerator $\sin (x^3-5x^2+ax+b)$ must also have a zero of order at least 2 at $x=2$.
This requires that the polynomial $P\left(x\right)=x^3-5x^2+ax+b$ has a root of multiplicity at least 2 at $x=2$.
Therefore:
1) $P\left(2\right)=0$
$2^3-5{\left(2\right)}^2+2a+b=0$
$8-20+2a+b=0\Rightarrow 2a+b=12$
2) $P^{\text{'}}\left(2\right)=0$
$P^{\text{'}}\left(x\right)=3x^2-10x+a$
$P^{\text{'}}\left(2\right)=3{\left(2\right)}^2-10\left(2\right)+a=0$
$12-20+a=0\Rightarrow a=8$
Substituting $a=8$ into the first condition:
$2\left(8\right)+b=12\Rightarrow b=-4$

Step 3: Factor the polynomial $P\left(x\right)$
With $a=8$ and $b=-4$, the polynomial is:
$P\left(x\right)=x^3-5x^2+8x-4$
Since $x=2$ is a double root, we can factor out ${(x-2)}^2=x^2-4x+4$:
$P\left(x\right)={(x-2)}^2(x-1)$

Step 4: Compute the limit $m$
Using the approximation $\sin \left(P\right(x\left)\right)\approx P\left(x\right)$ as $P\left(x\right)\to 0$:
$m=\underset{x\to 2}{lim}\frac{{(x-2)}^2(x-1)}{(\sqrt{x-1}-1){\log }_e(x-1)}$
Substitute $t=x-2$ where $t\to 0$:
$m=\underset{t\to 0}{lim}\frac{t^2(t+1)}{(\sqrt{t+1}-1){\log }_e(t+1)}$
Split the limit into separate standard limit forms:
$m=\underset{t\to 0}{lim}(t+1)·\left(\frac{t}{\sqrt{t+1}-1}\right)·\left(\frac{t}{{\log }_e(t+1)}\right)$
Evaluating each part as $t\to 0$:
- $\underset{t\to 0}{lim}(t+1)=1$
- $\underset{t\to 0}{lim}\frac{t}{\sqrt{t+1}-1}=\underset{t\to 0}{lim}\frac{t(\sqrt{t+1}+1)}{t}$
- $\underset{t\to 0}{lim}\frac{t}{{\log }_e(t+1)}=1$
Therefore:
$m=1·2·1=2$

Step 5: Calculate the value of $a+b+m$
We have:
$a=8$
$b=-4$
$m=2$
Substituting these values:
$a+b+m=8+(-4)+2=6$