Step 1: Understanding the Concept:
Two lines in 3D space are defined by their direction ratios (or cosines). The angle \(\theta\) between two lines with direction ratios \((l_1, m_1, n_1)\) and \((l_2, m_2, n_2)\) is given by:
\[ \cos\theta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}} \]
If the lines are perpendicular (\(\theta = \pi/2\)), the dot product of their direction ratios is zero: \(l_1l_2 + m_1m_2 + n_1n_2 = 0\).
Step 2: Detailed Explanation:
Given:
1) \(l + m + n = 0 \implies n = -(l + m)\).
2) \(mn - 2ln + lm = 0\).
Substitute the value of \(n\) from eq (1) into eq (2):
\[ m[-(l + m)] - 2l[-(l + m)] + lm = 0 \]
\[ -ml - m^2 + 2l^2 + 2lm + lm = 0 \]
\[ 2l^2 + 2lm - m^2 = 0 \]
Divide by \(m^2\) to form a quadratic in \((l/m)\):
\[ 2\left(\frac{l}{m}\right)^2 + 2\left(\frac{l}{m}\right) - 1 = 0 \]
Let the roots be \(x_1 = l_1/m_1\) and \(x_2 = l_2/m_2\).
Product of roots: \(\frac{l_1l_2}{m_1m_2} = \frac{-1}{2} \implies 2l_1l_2 = -m_1m_2 \implies 2l_1l_2 + m_1m_2 = 0\).
Now, from \(n = -(l+m)\), we find \(n_1n_2\):
\(n_1n_2 = (l_1+m_1)(l_2+m_2) = l_1l_2 + l_1m_2 + l_2m_1 + m_1m_2\).
From sum of roots \(x_1+x_2 = -1\): \(\frac{l_1}{m_1} + \frac{l_2}{m_2} = -1 \implies l_1m_2 + l_2m_1 = -m_1m_2\).
So, \(n_1n_2 = l_1l_2 - m_1m_2 + m_1m_2 = l_1l_2\).
Check the condition for perpendicularity:
\(l_1l_2 + m_1m_2 + n_1n_2 = l_1l_2 + m_1m_2 + l_1l_2 = 2l_1l_2 + m_1m_2\).
Since \(2l_1l_2 + m_1m_2 = 0\), the dot product is zero.
Thus, \(\cos\theta = 0 \implies \theta = \pi/2\).
Step 3: Final Answer:
The angle between the lines is \(\pi/2\).
Hence, the correct option is (C).