Consider the following $2$ stack PDA, it has two stacks, Stack $1$ and Stack $2.$

Pop$1$ means pop from stack $1$; pop$2$ means pop from stack $2$

Time left $1$:$58$:$01$

Push$1$ means push on top of stack $1$; Push$2$ means push on top of stack $2.$

a

a$.$ Adding one more stack to the PDA allows the PDA to accept languages of the form

$a^n{b}^n{c}^n$

b$.$ If you add a transition q$1-->$ q$3$: $,$pop$2($$$)$ push$2($$$)$ then the PDA will be able to accept the language $a^n{b}^n$

c$.$ For the string aaabbccc, the PDA will halt at state $q2,$ stack $1$ has one a$,$ and stack $2$ is empty and the string will be rejected

d$.$ for the PDA to accept $a^n{b}^n,$ state $q2$ must be made accepting state

e$.$ For the string aabbbccc

The PDA halts at q$1$ and the string is rejected, and stack $1$ is empty and stack $2$ has $b$ b