The equations of the straight line through the origin and parallel to the line (b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z are

$\frac{x}{a^2-bc}=\frac{y}{b^2-ca}=\frac{z}{c^2-ab}$

Equations of straight line through the origin are $\frac{x-0}{l}=\frac{y-0}{m}=\frac{z-0}{n}$

where l(b + c) + m(c + a) + n(a + b) = 0 and l(b - c) + m(c - a) + n(a - b) = 0.

On solving, $\frac{l}{2(a^2-bc)}\frac{m}{2(b^2-ca)}=\frac{n}{2(c^2-ab)}$

Equations of the straight line are $\frac{x}{a^2-bc}=\frac{y}{b^2-ca}=\frac{z}{c^2-ab}$.

Hence (C) is the correct answer.