Given points from which plane passing through are $(2,3,0),(0,-5,2)$ and $(-2,0,3)$.
Let the equation of plane be $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1$. Satisfy all the points,


$\frac{2}{a}+\frac{3}{b}=1$

Multiply equation (ii) by 3 and (iv) by 5 and then add,
$$
\begin{gathered}
\frac{-15}{b}+\frac{16}{c}=3 \\
\frac{\frac{15}{b}+\frac{15}{c}=10}{\frac{21}{c}=13}
\end{gathered}
$$
From (iii)
$$
\begin{aligned}
& \frac{-2}{\mathrm{a}}+\frac{3 \times 13}{21}=1 \\
& \Rightarrow \frac{13}{7}-1=\frac{2}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{7}{3}
\end{aligned}
$$
So, intercept of $\mathrm{A}$ is $(7 / 3,0,0)$