Column 6, when $\lim _{x \rightarrow-\infty}\left(4 x-\sqrt{a x^{2}-b x+c}=3\right.$,

 untranslated part: 

when $\lim _{x \rightarrow-\infty}\left(4 x-\sqrt{a x^{2}-b x+c}\right)=3$,
$$
\text { Let } \begin{aligned}
& \lim _{x \rightarrow \infty}\left(4 x-\sqrt{a x^{2}-b x+c}\right. \\
= & \lim _{x \rightarrow \infty}\left(4 x-\sqrt{a x^{2}-b x+c} \cdot \frac{4 x+\sqrt{a x^{2}-b x+c}}{4 x+\sqrt{a x^{2}-b x+c}}\right. \\
= & \frac{(16-a) x+b}{4+\sqrt{a}}
\end{aligned}
$$

Given the condition $\frac{(16-a) \mathbf{x}+\mathrm{b}}{4+\sqrt{a}}=3$,
$$
\therefore\left\{\begin{array}{c}
16-a=0 \\
b \\
4+\sqrt{a}
\end{array}\right.
$$

Solving it, we get $\mathrm{a}=16, \mathrm{~b}=24$.
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