A plane passing through the point $(-1,2,3)$, then its equation is
$$
a(x+1)+b(y-2)+c(z-3)=0
$$
where $\langle a, b, c\rangle$ are direction ratios of normal to the plane $A B C$.

So, the normal makes equal angles with coordinate axes i.e.,
$$
(a, b, c)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)
$$
Now, from Eq. (i),
$$
\begin{array}{cc}
\frac{1}{\sqrt{3}}(x+1)+\frac{1}{\sqrt{3}}(y-2)+\frac{1}{\sqrt{3}}(z-3)=0 \\
\Rightarrow \quad x+y+z-4 & =0
\end{array}
$$