Since the median of the integers is $6$, then $x \ge 6$.
Since the mean of the integers is $6$, then we have
$\begin{array}{rcl}
\dfrac{2+5+6+6+x+x+x+y}{8} & = & 6 \\
3x+y & 29
\end{array}$
Since $x$ and $y$ are positive integers and $x \ge 6$, then we have $\left\{ \begin{array}{l} x = 6 \\ y = 11 \end{array}\right.$ or $\left\{ \begin{array}{l} x = 7 \\ y = 8 \end{array}\right.$ or $\left\{ \begin{array}{l} x = 8 \\ y = 5 \end{array}\right.$ or $\left\{ \begin{array}{l} x = 9 \\ y = 2 \end{array}\right.$.
I may not be true. Assume $\left\{ \begin{array}{l} x = 7 \\ y = 8 \end{array}\right.$. Then the mode of the positive integers is $7$.
II must be true. For the case $\left\{ \begin{array}{l} x = 6 \\ y = 11 \end{array}\right.$, the range of the positive integers is $9$.
For the case $\left\{ \begin{array}{l} x = 7 \\ y = 8 \end{array}\right.$, the range of the positive integers is $6$.
For the case $\left\{ \begin{array}{l} x = 8 \\ y = 5 \end{array}\right.$, the range of the positive integers is $6$.
For the case $\left\{ \begin{array}{l} x = 9 \\ y = 2 \end{array}\right.$, the range of the positive integers is $7$.
III is not true. For the case $\left\{ \begin{array}{l} x = 9 \\ y = 2 \end{array}\right.$, the variance is $7.5$.
