$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$
of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^2=$
which gives $\alpha=-1$ or +1
For $\alpha=1$, the equations become
$ x+y+z=1$
$x+y+z=-1$
$and x+y+z=1$
which give no solution
For $\alpha=-1$, the equations become
$x-y+z=1$
$-x+y-z=-1$
$x-y+z=1$
which are all same and hence infinitely many solutions
Hence, $\alpha=-1$
$\Rightarrow 1+\alpha+\alpha^2=1$
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$\left\{( x , y ) \in R \times R : 0 \leq x \leq \frac{\pi}{2} \text { and } 0 \leq y \leq 2 \sin (2 x )\right\}$
and having one side on the $x$-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is