Here, the coordinates of the centre of mass of the three rods are:

\(\left(0, \frac{ L }{2}\right),\left(\frac{ L }{2}, 0\right),\left(\frac{ L }{2}, L \right)\)

The arrangement is symmetric about the line \(y=\frac{L}{2}\)

Hence the y coordinate of the centre of mass is \(\frac{L}{2}\).

Now,\(x\) coordinate of the centre of mass \(=\frac{\sum m_i x_i}{\sum m_i}\)

\(=\frac{ L }{3}\)

Therefore , the centre of mass of the arrangement is at \(\left(\frac{ L }{3}, \frac{ L }{2}\right)\).