If the 7th term in the binomial expansion of $$\left(\frac{3}{\sqrt[3]{84}} + \sqrt{3}\ln x\right)^9$$, $$x \gt 0$$, is equal to 729, then $$x$$ can be:

We begin with the binomial expansion formula. For any real numbers $$a$$ and $$b$$ and a positive integer $$n$$, the general term (also called the $$k^{\text{th}}$$ or $$(k+1)^{\text{th}}$$ term, depending on counting from zero) in $$(a+b)^n$$ is

$$T_{k+1}= {^nC_k}\, a^{\,n-k}\, b^{\,k},$$

where $$k = 0,1,2,\dots ,n$$ and $${^nC_k}=\dfrac{n!}{k!(n-k)!}$$.

In the given expression we have

$$a=\dfrac{3}{\sqrt[3]{84}},\qquad b=\sqrt{3}\,\ln x,\qquad n=9.$$

The problem refers to the 7th term, so we set $$k=6$$ (because $$T_{k+1}=T_7$$). Substituting these values into the general term formula, we get

$$T_7 = {^9C_6}\left(\dfrac{3}{\sqrt[3]{84}}\right)^{9-6}\bigl(\sqrt{3}\,\ln x\bigr)^{6}.$$

First, evaluate the binomial coefficient:

$$ {^9C_6}= \dfrac{9!}{6!\,3!}=84.$$

Next, handle the power of the first part:

$$\left(\dfrac{3}{\sqrt[3]{84}}\right)^{9-6}= \left(\dfrac{3}{\sqrt[3]{84}}\right)^{3} =\dfrac{3^{3}}{\bigl(\sqrt[3]{84}\bigr)^{3}} =\dfrac{27}{84} =\dfrac{9}{28}.$$

Putting these together, we obtain

$$T_7 = 84 \times \dfrac{9}{28} \times (\sqrt{3}\,\ln x)^{6}.$$

Simplify the numerical coefficient:

$$84 \times \dfrac{9}{28}= \bigl(\tfrac{84}{28}\bigr)\times 9 = 3 \times 9 = 27.$$

So

$$T_7 = 27\bigl(\sqrt{3}\,\ln x\bigr)^6.$$

According to the question, this 7th term equals $$729$$, hence

$$27\bigl(\sqrt{3}\,\ln x\bigr)^6 = 729.$$

Divide both sides by $$27$$:

$$(\sqrt{3}\,\ln x)^{6} = \dfrac{729}{27}=27.$$

Notice that $$27 = 3^{3}.$$ Therefore

$$(\sqrt{3}\,\ln x)^{6} = 3^{3}.$$

To isolate the inner expression, take the sixth root of both sides. Recall that $$3^{3} = 3^{\frac{6}{2}} = (3^{\frac12})^{6},$$ so

$$\sqrt{3}\,\ln x = 3^{\frac12} = \sqrt{3}.$$

Now divide by $$\sqrt{3}$$:

$$\ln x = 1.$$

Finally, use the definition of the natural logarithm ($$\ln x = 1 \Longrightarrow x = e^{1}$$):

$$x = e.$$

Looking at the provided options, $$e$$ corresponds to Option 2. Hence, the correct answer is Option 2.