The two lines \(lx + my = n\) and \(l'x + m'y = n'\) are perpendicular if the product of their slopes is -1.
First, let's find the slopes of the lines.
For the line \(lx + my = n\), we can rewrite it as \(my = -lx + n\), so \(y = -\frac{l}{m}x + \frac{n}{m}\). The slope, \(m_1\), is \(-\frac{l}{m}\).
For the line \(l'x + m'y = n'\), we can rewrite it as \(m'y = -l'x + n'\), so \(y = -\frac{l'}{m'}x + \frac{n'}{m'}\). The slope, \(m_2\), is \(-\frac{l'}{m'}\).
For the lines to be perpendicular, \(m_1 \cdot m_2 = -1\).
So, \((-\frac{l}{m}) \cdot (-\frac{l'}{m'}) = -1\)
\(\frac{ll'}{mm'} = -1\)
\(ll' = -mm'\)
\(ll' + mm' = 0\)
Therefore, the correct option is (A) \(ll' + mm' = 0\).