First off, I prove $2abc-ab-bc-ca$ is un-achievable. Also, assume WLOG $a\ge b\ge c$. Assume $2abc-bc-ca=xbc+yca+zab$, then take the equation $\pmod{ab}$ to give us $(x+1)bc+(y+1)ac\equiv 0\pmod{ab}$. By CRT and $\gcd(a,b)=1$, we take this equation mod a and mod b to give us $y+1\equiv 0\pmod{b}$, and $x+1\equiv 0\pmod{a}$ (and using all numbers are relatively prime pairwise). Substituting this back into the original equation gives us $z\le -1$, contradiction (this is where the $2abc$ part comes in). Now, let $2abc-bc-ca-ab+k=xbc+yca+zab$, and if a solution exists where $bc-1\ge k\ge 1$ then we are done because we can add $1$ to $x$ always. Consider the set $x=a-m$, where $m\in \{1,2,\cdots, a-1, a\}$ and $y=b-n$ where $n\in \{1,2,\cdots, b-1, b\}$. Therefore, we get $2abc-bc-ca-ab+k=(a-m)bc+(b-n)ca+zab$. This is the same as $bc(m-1)+ca(n-1)+k=(z+1)ab$ after doing some simplifying. By CRT, this must hold mod a and mod b and because $\gcd(a,b)=1$. Mod $a$ gives us $bc(m-1)+k\equiv 0\pmod{a}$, for which a value of $m-1$ obviously exists mod $a$, which can be chosen from the set of values we have assigned for $m$. Similar method shows a value of $n-1$ exists mod $b$ from the set we have given to $n$. Now, we already know that $x,y\ge 0$. Also, the LHS of the equation $bc(m-1)+ca(n-1)+k$ is at least $1$, therefore $z\ge 0$ and we are done.

First we will prove $2abc-ab-bc-ca$ is unattainable, as such: Suppose $xbc+yca+zab=2abc-ab-bc-ca$. Then, taking this mod $a$, we have that $xbc\equiv 0\pmod{a}$, so $x\equiv 0\pmod{a}$, and $a|x$. Similarly, $b|y$, and $c|z$, so $x\ge a$, $y\ge b$, and $z\ge c$. Thus, $xbc+yca+zac\ge 3abc$, so $2abc\ge 3abc$, which is a contradiction. Now we will prove all $n>2abc-ab-bc-ca$ is attainable, as such: consider the integer $r$ such that $r\equiv \frac{n}{bc} \pmod{a}$ and $0\le r\le a-1$. Rearranging the equation $xbc+yca+zab=n$ yields $\frac{n-xbc}{a}=yc+zb$, so set $x=r$. We see that $xbc-n=rbc-n\equiv 0 \pmod{a}$, so $\frac{n-xbc}{a}$ is a positive integer (obviously $n-xbc>0)$. Now, note that since $x\le a-1$, we have that $2abc-ab-ac-bc=(a-1)bc+abc-ab-ac\ge xbc+abc-ab-ac$, so $n>xbc+abc-ab-ac$. Thus, $n-xbc>abc-ab-ac$ so $\frac{n-xbc}{a}>bc-b-c$, so by Chicken Mc-Nugget theorem, there exist $b, c$ that satisfy the equation and are now done. $\blacksquare$

lemma - $bc - b - c$ is the largest number which cannot be written as $mb + nc$ with m and n non-negative Proof to lemma$0, c, 2c, ... , (b-1)c$ is a complete set of residues $mod b$ If $r > (b-1)c - b$, then $r = nc (mod b)$ for some $0 <= n <= b-1$ But $r > nc - b$, so $r = nc + mb$ for some $m => 0$ That proves that every number larger than $bc - b - c$ can be written as $mb + nc$ with $m$ and $n$ non-negative Now consider $bc - b - c$ It is $(b-1)c (mod b)$, and not congruent to any $nc$ where $0 <= n < b-1$ So if $bc - b - c = mb + nc$ then $n => b-1$ Hence $mb + nc => nc => (b-1)c > bc - b - c$ Contradiction. $0, bc, 2bc, ... , (a-1)bc$ is a complete set of residues $mod a$ So given $N > 2abc - ab - bc - ca$ we may take $xbc = N (mod a)$ with $0 <= x < a$ But $N - xbc > 2abc - ab - bc - ca - (a-1)bc = abc - ab - ca = a(bc - b - c)$ So $N - xbc = ka$, with $k > bc - b - c$ Hence we can find non-negative $y, z$ so that $k = zb + yc$ So, $N = xbc + yca + zab$ Finally, we show that for $N = 2abc - ab - bc - ca$ we cannot find non-negative $x, y, z$ so that $N = xbc + yca + zab$ $N = -bc (mod a)$ so we must have $x = -1 (mod a)$ and hence $x => a-1$ Similarly, $y => b-1, and z => c-1$ Hence $xbc + yca + zab => 3abc - ab - bc - ca > N$ Contradiction