Fun problem! This is an example of a trick you'll see from time to time on GMAT algebra problems. They want you to assume that, since there are 3 variables, you need at least 3 equations to solve for any one of them. However, in some cases (like this problem), multiple variables cancel out and it's possible to solve with only two equations.

I simplified the question first to isolate B, since that's what I'm trying to solve for:

2A + B = -3(2C - 1)
B = -6C + 3 - 2A
B = -2(A + 3C) + 3

Statement 1: Plug in the value given for A + 3C. (By the way, the GMAT won't use a coefficient of 1 in front of a variable like shown here.)

B = -2(-B/4) + 3
B = B/2 + 3
(1/2)B = 3
B = 6

This statement is sufficient.

If you didn't see that you can plug in the value for A + 3C - for example, if you didn't simplify the question in the same way as shown above - that's fine. Here's how it would play out:

2A + B = -3(2C - 1)
2A + B = -6C + 3

A + 3C = -B/4

Let's isolate A in the second equation, and then plug into the first one:

A = -3C - B/4

2(-3C - B/4) + B = -6C + 3
-6C - B/2 + B = -6C + 3
B/2 = 3
B = 6

Either way, you can solve for B and the statement is sufficient.

Statement 2: This is a good example of a "nice but not necessary" statement. It's useful info, but you don't need it to answer the question (and by itself, it doesn't help you answer the question.) We know that B = -2(A + 3C) + 3. So, B = -2(A - 3) + 3 = -2A + 6 + 3 = -2A + 9. However, without the value of A, we can't solve for B. This statement is insufficient.

The answer is A.
_________________

Since we need the value of \(b\), lets isolate \(b\)

On solving \(2a + b = −3(2c − 1)\), we get:
\(b =-6c-2a+3\) ( Lets call this main eqn.)
So to get the value of \(b\)we need the value of \(-6c-2a\)

(1) a + 3c = −\(\frac{1b}{4}\)
\(-b=4a+12c\)\(\hspace{1mm}\)or\(\hspace{2mm}\)\(b= -4a-12c\)\(\hspace{2mm}\)divide this eqn. by 2 we get\(\frac{b}{2}\) = \(-2a-6c\), so we have found \(-6c -2a\) in tems of \(b\), quick glance at the main eqn. above will tell us that this is SUFF. However if one wants to solve further then put \(-6c-2a\)=\(\frac{b}{2}\) in main eqn. we get
b=\(\frac{b}{2}\) +3 on solving this we will get \(b=6\) SUFF.

(2) \(c\) = −1, we cannot get the value of \(a\) hence the value of \(-6c-2a\) cannot be obtained. INSUFF.

Ans-A

Hope this helps.
_________________