Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3. This is the answer given:

I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start  with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.

Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.
$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.
The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizontal line and rotate back by angle $\gamma$)
\begin{align}T&=T_{-\gamma}.R_0.T_{\gamma}\\
&=\begin{pmatrix}cos \gamma & -sin\gamma\\ sin\gamma  & cos \gamma\end{pmatrix} 
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix} 
\begin{pmatrix}cos \gamma & sin\gamma\\ -sin\gamma  & cos \gamma\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} 
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix} 
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ -\frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} 
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & -\frac{m}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} \\
&=\begin{pmatrix}\frac{5}{13}&-\frac{12}{13}\\ -\frac{12}{13}&-\frac{5}{13}\end{pmatrix}
\end{align}
The following example in R shows how the black point (1,1) is reflected w.r.t. the life to new location (red point)
x <- seq(-10,10,0.01)
y <- -2*x/3
m <- -2/3
T <- matrix(c((1-m^2)/(1+m^2),2*m/(1+m^2),2*m/(1+m^2),(m^2-1)/(1+m^2)), nrow=2, byrow=T)
p <- c(1,2)
p1 <- T %*% p
plot(x,y, type='l')
points(p[1],p[2],pch=19)
points(p1[1],p1[2],pch=19, col='red')


