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(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.From $(1)$,$m = -2l - 2n$.Substitute $m$ into $(2)$: $3l^2 + 5(-2l

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$\frac{\pi}{2}$

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(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.From $(1)$,$m = -2l - 2n$.Substitute $m$ into $(2)$: $3l^2 + 5(-2l - 2n)^2 - 11n^2 = 0$.$3l^2 + 5(4l^2 + 8ln + 4n^2) - 11n^2 = 0$.$3l^2 + 20l^2 + 40ln + 20n^2 - 11n^2 = 0$.$23l^2 + 40ln + 9n^2 = 0$.Divide by $n^2$: $23(\frac{l}{n})^2 + 40(\frac{l}{n}) + 9 = 0$.Let $x = \frac{l}{n}$. Then $23x^2 + 40x + 9 = 0$.Let the roots be $x_1 = \frac{l_1}{n_1}$ and $x_2 = \frac{l_2}{n_2}$.Then $x_1 x_2 = \frac{l_1 l_2}{n_1 n_2} = \frac{9}{23}$.Similarly,substituting $l = -\frac{m+2n}{2}$ into $(2)$ gives $3(\frac{m+2n}{2})^2 + 5m^2 - 11n^2 = 0$,which leads to $3(m^2 + 4mn + 4n^2) + 20m^2 - 44n^2 = 0$,so $23m^2 + 12mn - 32n^2 = 0$.Dividing by $n^2$,$23(\frac{m}{n})^2 + 12(\frac{m}{n}) - 32 = 0$.Let $y_1 = \frac{m_1}{n_1}$ and $y_2 = \frac{m_2}{n_2}$. Then $y_1 y_2 = \frac{m_1 m_2}{n_1 n_2} = -\frac{32}{23}$.For two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,$\cos 	heta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.Since $l_i^2 + m_i^2 + n_i^2 = 1$,we use the relation $l_1 l_2 + m_1 m_2 + n_1 n_2 = n_1 n_2 (x_1 x_2 + y_1 y_2 + 1) = n_1 n_2 (\frac{9}{23} - \frac{32}{23} + 1) = n_1 n_2 (\frac{-23}{23} + 1) = 0$.Thus,$\cos 	heta = 0$,which means $	heta = \frac{\pi}{2}$.

If the direction cosines of two lines are such that $2l + m + 2n = 0$ and $3l^2 + 5m^2 - 11n^2 = 0$,then the angle between the two lines is

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