Value of ABC

adding all three eqs ..

a^2 + b^2 +c^2 + 6a +2b +4c = -14
a^2 + b^2 +c^2 = -14 - (6a +2b +4c) [even - even = even]

B, D and E are eliminated.

a^2 + 2 b = 7
odd + even = odd

a = odd

b^2 + 4 c = -7
odd + even = odd
b = odd

c^2 + 6 a = -14
even + even = even
c = even


a^2 + b^2 +c^2 = -14 - (6a +2b +4c)
dividing both sides by 2

(a^2 + b^2 +c^2)/2 = -7 - (3a +b +2c) = odd - (odd+odd-even) = odd


(a^2 + b^2 +c^2)/2 = odd

Only A satisfies.

Option A is the correct answer.

There may be a more sophisticated way to solve this, but the following worked for me.

First I solved for a^2, b^2 and c^2 and added them and got: -14 - 6a - 2b - 4c

That tells me that the answer must be even so it has to be a or c.

Since our answers are integers I assume the variables are integers.

Listing the perfect squares from 1 to 36 I get 1, 4, 9, 16, 25, and 36.
Our task is to find three of then that total to either 14 or 36.
1 + 4 + 9 = 14 so I chose that as my answer.

Checking the answer, (which you don't have to do on test day) we know that the absolute values of a, b, and c are 1, 2, and 3, not necessarily in that order. We also know from our original equations that at least c and a must be negative.
Playing with the possibilities I get a = -3, b = -1 and c = -2

Fooling around with the possibilitie

Add all of the equations together to get:

a^2+6a+b^2+2b+c^2+4c=-14

Complete the square for each variable:

a^2+6a+9 + b^2+2b+1 + c^2+4c+4 = -14+14

(a+3)^2 + (b+1)^2 + (c+2)^2 =0

The sum of three squares can only be zero if all three of them are zero, thus:

a=-3, b=-1. c=-2 and

a^2+b^2+c^2=9+1+4=14

Ans: A