If the system of linear equations
2x+2y+3z=a
3x-y+5z=b
x-3y+2z=c
where a,b and c are non-zero real numbers, has more than one solution, then

Options

A. b-c-a =0
B. a+b+c =0
C. b-c+a=0
D. b+c-a=0

Explanation

We know that, if the system of equations
`a_(1)x + b_(1)y + c_(1)z =d_(1)`
`a_(2)x + b_(2)y + c_(3)z = d_(2)`
`a_(3)x + b_(3)y + c_(3)z = d_(3)`
has more than one solution, then D =0 and `D_(1) = D_(2) = D_(3) =0`. In the given problem,
`D_(1) = 0 rArr |{:(a, 2, 3), (b, -1, 5), (c, -3, 2):}| =0`
`rArr a(-2+15)-2(2b-5c)+3(-3b+c)=0`
`rArr 13a + 4b + 10c -9b +3c =0`
`rArr 13a -13b +13c =0`
`rArr a-b+c = 0 rArr b-a-c =0`