# If (b+c)x+(c+a)y+(a+b)z=k=(b−c)x+(c−a)y+(a−b)z then what will be the equation of the straight line passing through origin and parallel to given line ?

If $\left(b+c\right)x+\left(c+a\right)y+\left(a+b\right)z=k=\left(b-c\right)x+\left(c-a\right)y+\left(a-b\right)z$ then what will be the equation of the straight line passing through origin and parallel to given line ?
I tried to relate the direction ratios of the given line and the new line as they are parallel but the problem is that line is passing through origin so I am ending up getting all zeros. Please help. Any hint will do .
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Kellen Blackburn
The given straight line is intersection of the planes ${\mathcal{P}}_{1}:\phantom{\rule{thickmathspace}{0ex}}\left(b+c\right)x+\left(c+a\right)y+\left(a+b\right)z=k$ and ${\mathcal{P}}_{2}:\phantom{\rule{thickmathspace}{0ex}}\left(b-c\right)x+\left(c-a\right)y+\left(a-b\right)z=k.$
Thanks to the coefficients we know that these planes are not parallel. Therefore, the parallel line passing through origin can be defined as intersection of planes parallel to ${\mathcal{P}}_{1},{\mathcal{P}}_{2}$ passing through origin. It is
$\left(b+c\right)x+\left(c+a\right)y+\left(a+b\right)z=0=\left(b-c\right)x+\left(c-a\right)y+\left(a-b\right)z.$
The vectors normal respectively to ${\mathcal{P}}_{1},{\mathcal{P}}_{2}$ represent diagonals of a parallelogram. The sides of this parallelogram are represented by the vectors (b,c,a) and (c,a,b). This is why the planes are not parallel.