$$
\begin{aligned}
& A=(a, 0,0), B \equiv(0, b, 0), C \equiv(0,0, c) \\
& \because \text { Centroid of } \triangle A B C \equiv\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \equiv(2,-3,5) \\
& \therefore a=6, b=-9, c=15
\end{aligned}
$$

Plane is $\frac{x}{6}-\frac{y}{9}+\frac{z}{15}=1$
$\therefore$ Perpendicular distance from origin is
$$
\begin{aligned}
\left|\frac{1}{\sqrt{\frac{1}{36}+\frac{1}{81}+\frac{1}{225}}}\right| & =\left|\frac{1}{\sqrt{\frac{81 \times 225+36 \times 225+36 \times 81}{225 \times 81 \times 36}}}\right| \\
& =\frac{15 \times 9 \times 6}{9 \sqrt{225+100+36}} \\
& =\left|\frac{90}{\sqrt{361}}\right|=\left|\frac{90}{19}\right|
\end{aligned}
$$