Ans: C
By intercepts form, we have
$\begin{array}{rcl}
\dfrac{x}{3} +\dfrac{\log_a y}{6} & = & 1 \\
2x+\log_a y& = & 6 \\
\log_a y & = & -2x +6 \\
y & = & a^{-2x+6} \\
y & = & a^6a^{-2x} \\
y & = & a^6 (a^{-2})^x
\end{array}$
By comparing the terms, we have $m=a^6$ and $n=a^{-2}$.
I is true. Since $0< a<1$, $0< a^6 <1$. Then $m< 1$.
II is not true. Since $0< a <1$, $a^{-2} >1$. Then $n>1$.
III is true. For $x=3$, we have
$\begin{array}{rcl}
y & = & a^6 (a^{-2})^3 \\
y & = & a^6 \times a^{-6} \\
y & = & a^{6-6} \\
y & = & a^0 \\
y & = & 1
\end{array}$
