230. The edge length of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is $a$. Points $E$ and $F$ are the midpoints of edges $B B_{1}$ and $C C_{1}$, respectively. Triangles are considered, whose vertices are the points of intersection of a plane parallel to the plane $A B C D$ with the lines $A C_{1}, C E$ and $D F$. Find the minimum value of the area of the considered triangles.
230. Consider $\triangle K L M$, representing the section of the given triangle on the plane $A B C D, K$ - on the line $C B$, $L$ - on $C D, M$ - on $C A$. If $|C K|=x$, then $|C L|=|a-x|$, $|C M|=\sqrt{\overline{2}}\left|a-\frac{x}{2}\right|$

It is not difficult to obtain that

$$
S_{K L M}=\frac{1}{2}\left|x(a-x)-a\left(a-\frac{x}{2}\right)\right|=\frac{1}{4}\left(2 x^{2}-3 a x+2 a^{2}\right)
$$

The minimum value is $\frac{7 a^{2}}{32}$.