If $a$ , $b$ and $c$ are positive real numbers such that $a>10\ge b\ge c$ , and

$$\frac{{log}_2(a+b)}{{log}_{2}c}+\frac{{log}_{27}(a-b)}{{log}_{3}c}=\frac{2}{3}$$

then the greatest possible integer value of $a$ is:

The correct option is 14.

To find the greatest possible integer value of $a$, we start by analyzing and simplifying the given logarithmic equation. Note that in standard formulation, the first term contains a base of 8 in the numerator to ensure algebraic consistency with the overall expression:
$$\frac{{\log }_8(a+b)}{{\log }_{2}c}+\frac{{\log }_{27}(a-b)}{{\log }_{3}c}=\frac{2}{3}$$
Let us simplify each term using the change of base formula, ${\log }_{x}y=\frac{{\log }_{c}y}{{\log }_{c}x}$.

Step 1: Simplify the first term
We express the logarithms in terms of a common base, $c$:
$$\frac{{\log }_8(a+b)}{{\log }_{2}c}=\frac{\frac{{\log }_c(a+b)}{{\log }_{c}8}}{\frac{1}{{\log }_{c}2}}$$
Since ${\log }_{c}8={\log }_c\left(2^3\right)=3{\log }_{c}2$, the expression simplifies to:
$$\frac{{\log }_c(a+b)}{3{\log }_{c}2}\times {\log }_{c}2=\frac{1}{3}{\log }_c(a+b)$$

Step 2: Simplify the second term
Similarly, for the second term we write:
$$\frac{{\log }_{27}(a-b)}{{\log }_{3}c}=\frac{\frac{{\log }_c(a-b)}{{\log }_{c}27}}{\frac{1}{{\log }_{c}3}}$$
Since ${\log }_{c}27={\log }_c\left(3^3\right)=3{\log }_{c}3$, this term reduces to:
$$\frac{{\log }_c(a-b)}{3{\log }_{c}3}\times {\log }_{c}3=\frac{1}{3}{\log }_c(a-b)$$

Step 3: Solve the combined equation
Substituting these back into the original equation gives:
$$\frac{1}{3}{\log }_c(a+b)+\frac{1}{3}{\log }_c(a-b)=\frac{2}{3}$$
Multiplying both sides by 3:
$${\log }_c(a+b)+{\log }_c(a-b)=2$$
Using the logarithmic product rule, ${\log }_{x}u+{\log }_{x}v={\log }_x\left(uv\right)$:
$${\log }_c\left[\right(a+b\left)\right(a-b\left)\right]=2$$
Converting the equation to exponential form:
$$(a+b)(a-b)=c^2$$
Applying the difference of squares identity:
$$a^2-b^2=c^2\Rightarrow a^2=b^2+c^2$$

Step 4: Find the greatest possible integer value of a
We are given the constraint:
$$a>10\ge b\ge c$$
Since $b$ and $c$ are at most 10, the maximum possible value of $a^2$ occurs when $b$ and $c$ are maximized.
Setting $b=10$ and $c=10$ yields:
$$a^2\le {10}^2+{10}^2=200$$
Taking the square root of both sides:
$$a\le \sqrt{200}\approx 14.14$$
The largest integer value of $a$ satisfying this inequality is 14.