6. Given that $a$ is a real number, and for any $k \in$ $[-1,1]$, when $x \in(0,6]$, we have
$$
6 \ln x+x^{2}-8 x+a \leqslant k x \text {. }
$$

Then the maximum value of $a$ is $\qquad$ .
$6.6-6 \ln 6$.
By the problem, for $k \in[-1,1]$,
$$
k \geqslant \frac{6 \ln x}{x}+\frac{a}{x}+x-8
$$

holds for all $x \in(0,6]$, then
$$
-1 \geqslant \frac{6 \ln x}{x}+\frac{a}{x}+x-8
$$

holds for all $x \in(0,6]$, which means
$$
a \leqslant-x^{2}-6 \ln x+7 x
$$

holds for all $x \in(0,6]$.
Let $h(x)=-x^{2}-6 \ln x+7 x(x \in(0,6])$.
Then $a_{\text {max }}=h_{\text {min }}(x)$.
So $h^{\prime}(x)=-2 x-\frac{6}{x}+7=\frac{-(2 x-3)(x-2)}{x}$.
Since $x>0$, therefore,
when $x \in\left(\frac{3}{2}, 2\right)$, $h^{\prime}(x)>0, h(x)$ is an increasing function;
when $x \in\left(0, \frac{3}{2}\right)$ and $x \in(2,+\infty)$, $h^{\prime}(x)$ $0,
\end{array}
$$

Thus, $h_{\min }(x)=h(6)=6-6 \ln 6$.
Therefore, the maximum value of $a$ is $6-6 \ln 6$.