The distance $(d)$ of a point $M\left(x_{0}, y_{0}\right)$ from the line $a x+b y+c=0$ is given by

$$
d=\frac{a x_{0}+b y_{0}+c}{ \pm \sqrt{a^{2}+b^{2}}}
$$

where $\sqrt{a^{2}+b^{2}}$ should be taken with the sign that makes $\frac{c}{ \pm \sqrt{a^{2}+b^{2}}}$ always negative.

If the radius $r$ and the coordinates of the center $(x_{0}, y_{0})$ of the circle inscribed in the triangle formed by the given lines are known, then

$$
\begin{gathered}
\frac{y_{0}+5}{-1}=r \\
\frac{3 x_{0}+4 y_{0}-28}{5}=r \\
\frac{-5 x_{0}+2 y_{0} \sqrt{6}-24}{7}=r
\end{gathered}
$$

thus

$$
\begin{gathered}
r+y_{0}=-5 \\
5 r-3 x_{0}-4 y_{0}=-28 \\
7 r+5 x_{0}-2 \sqrt{6} y_{0}=-24
\end{gathered}
$$

from which

$$
r=\left|\begin{array}{ccc}
-5 & 0 & 1 \\
-28 & -3 & -4 \\
-24 & 5 & -2 \sqrt{6}
\end{array}\right|:\left|\begin{array}{ccc}
1 & 0 & 1 \\
5 & -3 & -4 \\
7 & 5 & -2 \sqrt{6}
\end{array}\right|=\frac{-1}{115}(542+3 \sqrt{6})
$$

If we do not consider the sign of the distance, then the desired diameter is

$$
d=2 r=\frac{2}{115}(542+\sqrt{6)}=9.55
$$

(Adolf Szücs, Budapest.)

The problem was also solved by: Ádámffy E., Bartók I., Dömény I., Haar A., Kertész G., Kürti I., Rássy P., Rosenberg J., Schöffer J., Schwemmer J., Szűcs A., Tóth B.
Given the equations of three lines:

$$
y+5=0, \quad 3 x+4 y=28, \quad 2 y \sqrt{6}-5 x=24
$$

find the diameter of the circle inscribed in the triangle formed by these three lines.