Which of the following statements is/are correct?
A. If $$2^{x}=3^{y}=6^{-z}, then \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$
B. $$(243)^{0.16}\times(9)^{0.1}=0.3$$
C. If $$3.105 \times10^{p}=0.00239 + 0.000715$$, then P = -3

Let's analyse the statements:

Statement A:

Let, $$2^{x}=3^{y}=6^{-z}=k$$

Now, $$2^x=k$$

or, $$x=\dfrac{\log\ k}{\log\ 2}$$

or, $$\dfrac{1}{x}=\dfrac{\log\ 2}{\log\ k}$$

Also, $$3^y=k$$ 

or, $$y=\dfrac{\log\ k}{\log\ 3}$$

or, $$\dfrac{1}{y}=\dfrac{\log\ 3}{\log\ k}$$

Also,$$6^-z=k$$

or, $$z=-\dfrac{\log\ k}{\log\ 6}$$

or, $$\dfrac{1}{z}=-\dfrac{\log\ k}{\log\ 6}$$

So, $$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{\log2}{\log k}+\dfrac{\log3}{\log k}-\dfrac{\log6}{\log k}=\dfrac{\log2+\log3-\log6}{\log\ k}=\dfrac{\log\left(2\times\ 3\right)-\log6}{k}=\dfrac{\log6-\log6}{k}=0$$

So, statement A is true

Statement B:

$$(243)^{0.16}\times(9)^{0.1}$$

=$$\left(3^5\right)^{0.16}\times\ \left(3^2\right)^{0.1}=3^{0.8}\times\ 3^{0.2}=3^{0.8+0.2}=3^1=3$$

But according to statement B the value must come 0.3

So, statement B is false.

Statement C:

$$3.105 \times10^{p}=0.00239 + 0.000715$$

or, $$3.105 \times10^{p}=0.003105$$

or, $$10^p=\dfrac{0.003105}{3.105}$$

or, $$10^p=\dfrac{1}{1000}=10^{-3}$$

So, p=-3

So, statement C is true.

So, option D is the correct