Identification of Formal Language
$L$ is CFL because there is a CFG generating it:

$$
\begin{align}
S&amp;\rightarrow aAc\\
A&amp;\rightarrow aAc \mid BC\\
B&amp;\rightarrow aDb\\
C&amp;\rightarrow bEc\\
D&amp;\rightarrow aDb \mid \epsilon\\
E&amp;\rightarrow bEc \mid \epsilon
\end{align}
$$



$L$ is not DCFL. To prove this, we give some lemmas firstly:


  Lemma 1. 
  $$
\begin{align}
L &amp;=\{a^x b^y c^z\mid x,y\ge 2, x+y+z\text{ is even},|x-y|+2\le z\le x+y-2\}\\
&amp;=\{a^x b^y c^{|x-y|+2+z}\mid x,y\ge 2, z\text{ is even},0\le z\le x+y-|x-y|-4\}.
\end{align}
$$


Proof. Let $L'=\{a^x b^y c^z\mid x,y\ge 2, x+y+z\text{ is even},|x-y|+2\le z\le x+y-2\}$. Easy to see $L\subseteq L'$. Now let $a^xb^yc^z\in L'$, then we can choose $m=(x+y-z)/2, n=(x-y+z)/2, k=(y-x+z)/2$. Because $x+y+z$ is even, $m,n,k$ are all integers. Because $|x-y|+2\le z\le x+y-2$, we have $m,n,k\ge 1$, so $a^xb^yc^z\in L$. As a result, $L=L'$.


  Lemma 2. $M =\{a^x b^y c^{|x-y|+2}d^z\mid x,y\ge 2, z\text{ is even},z\le x+y-|x-y|-4\}$ is not context-free.


Proof. Suppose it is context-free, according to Odgen's lemma, there exists some $p\ge 1$ such that $s=a^{p+2}b^{p+2}c^2d^{2p}\in M$ can be written as $s=uvwrt$, such that


the number of $d$s in $vwr$ is at most $p$ (i.e. we mark all $d$s),
there is at least one $d$ in $vr$, and
for all $q\ge 0$, $uv^qwr^qt\in M$.


From condition 3, $v$ and $r$ each contains at most one character. Together with condition 2, as $q$ grows, the number of $d$s in $uv^qwr^qt$ grows infinitely. To satisfy the condition $z\le x+y-|x-y|-4=2\min\{x,y\}-4$, the number of $a$ and $b$ must both grow, i.e. $a$ and $b$ must both in $vr$. However, $vr$ contains at most two characters, while $d$ is already in it, $vr$ cannot contain both $a$ and $b$, a contradiction.



Now let's come back to $L$. Suppose there is a DPDA $D$ accepting $L$, we create two copies $D_1$ and $D_2$ of $D$ and change the input character $c$ in $D_2$ to $d$. We then construct a new PDA $P$ as follows:


The states of $P$ are the union of states of $D_1$ and $D_2$. The start state of $P$ is the start state of $D_1$. The accepting states of $P$ are the accepting states of $D_2$.
For a transition in $D_1$, if the destination is an accepting state in $D_1$, change the destination to the corresponding state in $D_2$. Other transitions keep unchanged.


Now run $P$ on a string $a^xb^yc^{|x-y|+2}d^z\in M$. According to lemma 1, after reading $a^xb^yc^{|x-y|+2}$, it enters a state in $D_2$ for the first time. Since $a^xb^yc^{|x-y|+2+z}\in L$ according to lemma 1, $P$ will finally accept.

On the other hand, if a string $s$ is accepted by $P$, let $s=s_1s_2$ where after reading $s_1$, $P$ enters a state in $D_2$ for the first time. According to lemma 1, $s_1$ must have the form $a^xb^yc^{|x-y|+2}$ ($x,y\ge 2$). Also, $s_2$ does not contain $c$, and after changing all $d$s in $s_2$ to $c$s, $s_1s_2$ belongs to $L$. This means $s$ must have the form $a^xb^yc^{|x-y|+2}d^z$ where $x+y+|x-y|+2+z$ is even (i.e. $z$ is even) and $|x-y|+2+z\le x+y-2$ (i.e. $z\le x+y-|x-y|-4$). Hence $s\in M$.

As a result, $P$ recognizes $M$, which contradicts to lemma 2.