The angle between the lines whose direction ratios are $a, b, c$ and $b - c, c - a, a - b$ is

Let the direction ratios of the two lines be represented by vectors $\vec{d_1}$ and $\vec{d_2}$.

$\vec{d_1} = a\hat{i} + b\hat{j} + c\hat{k}$

$\vec{d_2} = (b - c)\hat{i} + (c - a)\hat{j} + (a - b)\hat{k}$

The angle $\theta$ between the lines is given by $\cos\theta = \dfrac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$.

Calculating the dot product of the direction vectors:

$\vec{d_1} \cdot \vec{d_2} = a(b - c) + b(c - a) + c(a - b)$

$\vec{d_1} \cdot \vec{d_2} = ab - ac + bc - ab + ac - bc = 0$

Since the dot product is zero, the vectors are perpendicular.

$\cos\theta = 0 \Rightarrow \theta = 90^\circ$

Answer: $90^\circ$