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 ?
My attempt:
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 .
The given straight line is intersection of the planes $ \mathcal{P}_1:\;(b+c)x+(c+a)y+(a+b)z=k$ and $\mathcal{P}_2:\;(b-c)x+(c-a)y+(a-b)z=k.$
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 $$(b+c)x+(c+a)y+(a+b)z=0=(b-c)x+(c-a)y+(a-b)z.$$
Note added
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.