The angle between lines whose direction cosines satisfy the equation $\ell+m+n=0$ and $\ell^2-m^2-n^2=0$, is

The direction cosines $(\ell, m, n)$ satisfy $\ell + m + n = 0$ and $\ell^2 - m^2 - n^2 = 0$.

From the first equation, $n = -(\ell + m)$. Substituting into the second equation:

$\ell^2 - m^2 - (\ell+m)^2 = 0 \implies -2m^2 - 2\ell m = 0 \implies -2m(m+\ell) = 0$

Two cases emerge from this equation.

Case 1: $m = 0$

From $\ell + m + n = 0$ and $m = 0$, we find $\ell = -n$. Using the identity $\ell^2 + m^2 + n^2 = 1$ yields $2n^2 = 1$, so $n = \pm \frac{1}{\sqrt{2}}$.
One solution gives direction cosines $\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$.

Case 2: $\ell = -m$

From $\ell + m + n = 0$ and $\ell = -m$, we find $n = 0$. The identity becomes $2m^2 = 1$, so $m = \pm \frac{1}{\sqrt{2}}$.
One solution gives direction cosines $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$.

The angle $\theta$ between these lines is given by:

$\cos \theta = |\ell_1 \ell_2 + m_1 m_2 + n_1 n_2| = \left| \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) + 0\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\cdot 0 \right| = \frac{1}{2}$

Therefore, $\theta = rccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

Final answer: $\boxed{\text{B}}$