`(a+2b)x+(2a-b)y+a+2b = 0`
`=>ax+2bx+2ay-by+a+2b= 0`
`=>a(x+2y+1)+b(2x-y+2)= 0`
`=>(x+2y+1)+b/a(2x-y+2) = 0`
Above equation represents equation of family of lines with `lambda = b/a`.
So, we can find intersection of these two lines.
`x+2y+1 = 0->(1)`
`2x-y+2 = 0->(2)`
Multiplying (2) by 2 and then adding (1) and (2) from it,
`x+2y+1 +4x-2y+4 = 0`
`=>5x = -5=> x = -1`
Putting value of ` x= -1` in (1),
`-1+2y+1 = 0=> y = 0`
So, if we verify point (-1,0) on the given option, we find, all of the option satisfies it.
But, we have to check for equal non-zero intercepts. So, option A and option C are correct options.
