The equation of the straight line passing through $R$ and $Q$ is
$\begin{array}{rcl}
\dfrac{x}{24} + \dfrac{y}{24} & = & 1 \\
x + y – 24 & = & 0
\end{array}$
Therefore, the coordinates of $R$ are $(12, 12)$ and the coordinates of $Q$ are $(18,6)$.
The equation of the straight line passing through $S$ and $P$ is
$\begin{array}{rcl}
\dfrac{x}{12} + \dfrac{y}{24} & = & 1 \\
2x + y – 24 & = & 0
\end{array}$
Therefore, the coordinates of $S$ are $(6,12)$ and the coordinates of $P$ are $(9,6)$.
At point $P$, the value of $7y-5x+3$
$\begin{array}{cl}
= & 7(6) -5(9) + 3 \\
= & 0
\end{array}$
At point $Q$, the value of $7y-5x+3$
$\begin{array}{cl}
= & 7(6) -5(18) + 3 \\
= & -45
\end{array}$
At point $R$, the value of $7y-5x+3$
$\begin{array}{cl}
= & 7(12) -5(12) + 3 \\
= & 27
\end{array}$
At point $S$, the value of $7y-5x+3$
$\begin{array}{cl}
= & 7(12) -5(6) + 3 \\
= & 57
\end{array}$
Hence, the value of $7y-5x+3$ attains its maximum at point $S$.
