1. Use the reflection property on the $x$-axis

A ray comes from $P(1,2)$, reflects at a point $Q$ on the $x$-axis, and then passes through $R(4,3)$.

For reflection from the $x$-axis, reflect the point $R(4,3)$ across the $x$-axis to $R'(4,-3).$ Then the reflected path $P \to Q \to R$ is equivalent to the straight line path $P \to Q \to R'.$ So $Q$ is the point where the line joining $P(1,2)$ and $R'(4,-3)$ meets the $x$-axis.

2. Find the equation of line $PR'$

Slope of the line through $P(1,2)$ and $R'(4,-3)$ is $m=\frac{-3-2}{4-1}=\frac{-5}{3}.$

Equation through $P(1,2)$: $y-2=-\frac{5}{3}(x-1).$

Since $Q$ lies on the $x$-axis, $y=0$. Hence $0-2=-\frac{5}{3}(x-1).$ So, $-2=-\frac{5}{3}(x-1)$ $2=\frac{5}{3}(x-1)$ $x-1=\frac{6}{5}$ $x=\frac{11}{5}.$

Therefore, $Q\left(\frac{11}{5},0\right).$

3. Use the parallelogram condition

If $PQRS$ is a parallelogram with vertices in order, then diagonals bisect each other, or equivalently $\vec{P}+\vec{R}=\vec{Q}+\vec{S}.$ Thus, $S=P+R-Q.$

Now, $P=(1,2),\quad Q=\left(\frac{11}{5},0\right),\quad R=(4,3).$ So