If $$\left(\frac{3^6}{4^4}\right) k$$ is the term, independent of $$x$$, in the binomial expansion of $$\left(\frac{x}{4} - \frac{12}{x^2}\right)^{12}$$, then $$k$$ is equal to _________.

For the binomial

$$\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$$

the general term (using the Binomial Theorem $$\bigl(a+b\bigr)^n=\displaystyle\sum_{r=0}^{n}{^nC_r}\,a^{\,n-r}\,b^{\,r}$$) is

$$T_{r+1}={^{12}C_{r}}\left(\frac{x}{4}\right)^{\,12-r}\!\!\left(-\frac{12}{x^{2}}\right)^{r}.$$

First we collect the powers of $$x$$ present in this term.

From the first factor we get $$x^{\,12-r}$$ and from the second factor $$x^{-2r}$$. Therefore the combined power of $$x$$ is

$$x^{\,(12-r)+(-2r)}=x^{\,12-3r}.$$

To obtain the term independent of $$x$$ we must set the exponent of $$x$$ equal to $$0$$:

$$12-3r=0\quad\Longrightarrow\quad r=4.$$

Now we substitute $$r=4$$ back into the expression for the general term.

$$T_{5}={^{12}C_{4}}\left(\frac{x}{4}\right)^{\,8}\!\!\left(-\frac{12}{x^{2}}\right)^{4}.$$

Simplifying step by step:

$$\begin{aligned}

T_{5}&={^{12}C_{4}}\;\frac{x^{8}}{4^{8}}\;(-1)^{4}\;\frac{12^{4}}{x^{8}} \\[4pt]

&={^{12}C_{4}}\;\frac{12^{4}}{4^{8}}\;x^{8-8} \\[4pt]

&={^{12}C_{4}}\;\frac{12^{4}}{4^{8}},\qquad(\text{since the powers of }x\text{ cancel}).

\end{aligned}$$

Because $$12=3\cdot4$$, we rewrite $$12^{4}=(3\cdot4)^{4}=3^{4}\,4^{4}$$. Substituting this gives

$$T_{5}={^{12}C_{4}}\;\frac{3^{4}\,4^{4}}{4^{8}}

={^{12}C_{4}}\;\frac{3^{4}}{4^{4}}.$$

Now we put actual numerical values.

$$^{12}C_{4}=\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2\cdot1}=495,\qquad 3^{4}=81,\qquad 4^{4}=256.$$

Thus

$$T_{5}=495\;\frac{81}{256}=\frac{40\,095}{256}.$$

Observe that

$$40\,095=55\times729\quad\text{and}\quad729=3^{6}.$$

Hence we can write

$$T_{5}=55\;\frac{3^{6}}{4^{4}}.$$

The problem states that the term independent of $$x$$ equals

$$\left(\frac{3^{6}}{4^{4}}\right)k.$$

Therefore,

$$k=55.$$

So, the answer is $$55$$.