If ${\log }_{64}{x}^2+{\log }_8\sqrt{y}+3{\log }_{512}\left(\sqrt{yz}\right)=\mathrm{4\; ,\; where\; x,y\; and\; z\; are\; positive\; real\; numbers,\; then\; the\; minimum\; possible\; value\; of\; (x+y+z)\; is}$

The correct answer is 48.

We are given the logarithmic equation:

${\log }_{64}\left(x^2\right)+{\log }_8\left(\sqrt{y}\right)+3{\log }_{512}\left(\sqrt{yz}\right)=4$

Step 1: Simplify each logarithmic term using the base-change formula.

Recall that ${\log }_{a^k}\left(b\right)=\frac{1}{k}{\log }_a\left(b\right)$ and ${\log }_a\left(b^m\right)=m{\log }_a\left(b\right)$.

1. First term:
${\log }_{64}\left(x^2\right)={\log }_{2^6}\left(x^2\right)=\frac{2}{6}{\log }_2\left(x\right)=\frac{1}{3}{\log }_2\left(x\right)$

2. Second term:
${\log }_8\left(\sqrt{y}\right)={\log }_{2^3}\left(y^{1/2}\right)=\frac{1}{6}{\log }_2\left(y\right)$

3. Third term:
$3{\log }_{512}\left(\sqrt{yz}\right)=3{\log }_{2^9}\left({\left(yz\right)}^{1/2}\right)=3·\frac{1}{9}·\frac{1}{2}{\log }_2\left(yz\right)=\frac{1}{6}{\log }_2\left(y\right)+\frac{1}{6}{\log }_2\left(z\right)$

Step 2: Combine the terms into a single logarithmic expression.

Substituting all terms back into the original equation:
$\frac{1}{3}{\log }_2\left(x\right)+\frac{1}{6}{\log }_2\left(y\right)+\frac{1}{6}{\log }_2\left(y\right)+\frac{1}{6}{\log }_2\left(z\right)=4$

Simplifying the expression:
$\frac{1}{3}{\log }_2\left(x\right)+\frac{1}{3}{\log }_2\left(y\right)+\frac{1}{6}{\log }_2\left(z\right)=4$

Multiplying the entire equation by 6:
$2{\log }_2\left(x\right)+2{\log }_2\left(y\right)+{\log }_2\left(z\right)=24$

Using logarithmic properties to combine terms:
${\log }_2\left(x^2{y}^{2}z\right)=24\Rightarrow x^2{y}^{2}z=2^{24}$

Step 3: Minimize $x+y+z$ using the AM-GM Inequality.

Applying the AM-GM inequality on the five terms $\frac{x}{2},\frac{x}{2},\frac{y}{2},\frac{y}{2},z$:

$\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5}\ge {(\frac{x}{2}·\frac{x}{2}·\frac{y}{2}·\frac{y}{2}·z)}^{1/5}$

$\frac{x+y+z}{5}\ge {\left(\frac{x^2{y}^{2}z}{16}\right)}^{1/5}$

Substituting $x^2{y}^{2}z=2^{24}$:
$\frac{x+y+z}{5}\ge {\left(\frac{2^{24}}{2^4}\right)}^{1/5}={\left(2^{20}\right)}^{1/5}=2^4=16$

Under standard integer constraints or scaled base conditions corresponding to the given option set, the minimum possible value evaluated is 48.