Is there a way to obtain exactly 2 quarts in the 8-quart or 5-quart pitcher? Suppose we are given pitchers of waters, of sizes $12$ quarts, $8$ quarts, and $5$ quarts. Initially the $12$ quart pitcher is full and the other two empty. We can pour water from one pitcher to another, pouring until the receiving pitcher is full or the pouring pitcher is empty. Is there a way to pour among pitchers to obtain exactly $2$ quarts in the $8$-quart or $5$-quart pitcher? If so, find the minimal sequence of pourings to get $2$ quarts in the $8$-quart or the $5$-quart pitcher. 
My work: 
Start at $(12,0,0)$
Can either go to $(7,0,5)$ or $(4,8,0)$
From $(7,0,5)$ you can go to $(0,7,5)$ or $(7,5,0)$
From $(4,8,0)$ you can go to $(0,8,4)$ or $(4,3,5)$
After this I'm kind of confused as to where to go with this. The answer in the back of the textbook says (listed as $(b,c)$) -> $(0,0)-(0,5)-(7,5)$. This answer doesn't make any sense to me and I don't know if it is correct. Am I missing something?
Edit: New thoughts are $(12,0,0)$ to $(7,0,5)$ to $(0,7,5)$ to $(5,7,0)$ to $(5,2,5)$ Is this the shortest way to solve this problem?

I would go to (7, 0, 5), then (7, 5, 0), (2, 5, 5) and (2, 8, 2).  This is four pours.
