Consider the simultaneous equations
$\left\{ \begin{array}{ll}
x+3=0 & \ldots \unicode{x2460} \\
2x+3y-12=0 & \ldots \unicode{x2461} \\
5x-3y+12=0 & \ldots \unicode{x2462}
\end{array}\right.$
From $\unicode{x2460}$, we have $x=-3$.
Sub. $x=-3$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
2(-3)+3y-12 & = & 0 \\
3y & = & 18 \\
y & = & 6
\end{array}$
Therefore, the coordinates of the intersection point of $\unicode{x2460}$ and $\unicode{x2461}$ are $(-3,6)$.
Sub. $x=-3$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
5(-3)-3y+12 & = & 0 \\
-3y & = & 3 \\
y & = & -1
\end{array}$
Therefore, the coordinates of the intersection point of $\unicode{x2460}$ and $\unicode{x2462}$ are $(-3,-1)$.
$\unicode{x2461} +\unicode{x2462}$, we have
$\begin{array}{rcl}
7x & = & 0 \\
x & = & 0
\end{array}$
Sub. $x=0$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
2(0)+3y-12 & = & 0 \\
3y & = & 12 \\
y & = & 4
\end{array}$
Therefore, the coordinates of the intersection point of $\unicode{x2461}$ and $\unicode{x2462}$ are $(0,4)$.
Note that the greatest value of $\beta x+6y$ is $24$.
For the point $(-3,6)$, we have
$\begin{array}{rcl}
\beta(-3)+6(6) & \le & 24 \\
-3\beta +36 & \le & 24 \\
-3\beta & \le & -12 \\
\beta & \ge & 4
\end{array}$
For the point $(-3,-1)$, we have
$\begin{array}{rcl}
\beta(-3)+6(-1) & \le & 24 \\
-3\beta-6 & \le & 24 \\
-3\beta & \le & 30 \\
\beta & \ge & 10
\end{array}$
For the point $(0,4)$, we have
$\begin{array}{rcl}
\beta(0)+6(4) & \le & 24 \\
24 & \le & 24
\end{array}$
Therefore, $\beta \ge 4$ and $\beta \ge -10$ and $24\le 24$.
Hence, the overall solution is $\beta \ge 4$.
