$B=\{0^n\#0^{2n}\#0^{3n}\mid n\geq0\}$
We use the pumping lemma for CFLs.
A. True
Choose $s=0^p\#0^{2p}\#0^{3p}$, where $p$ is the pumping length.
Clearly, $|s|\geq p$.
B. True
For a CFL pumping decomposition,
$s=uvxyz$ with $|vxy|\leq p$ and $|vy|>0$.
If $v$ or $y$ contains a $#$, then pumping with $i=0$ or $i=2$ changes the number or arrangement of $\#$ symbols.
But every string in $B$ must contain exactly two $\#$ symbols in the required positions.
So the pumped string is not in $B$.
C. True
Suppose neither $v$ nor $y$ contains $\#$.
Since $|vxy|\leq p$, the substring $vxy$ cannot cover all three zero-blocks.
Thus pumping can change only one block or at most two adjacent blocks.
But a string in $B$ must have block lengths in the ratio $1:2:3$.
Pumping changes some block lengths while leaving at least one block unchanged, so this ratio cannot remain $1:2:3$.
Therefore, for some pumping value such as $i=0$ or $i=2$, the resulting string is not in $B$.
D. False
One stack cannot, in general, enforce all three conditions
$|B_2|=2|B_1|$ and $|B_3|=3|B_1|$
while preserving the required information.
Indeed, $B$ is not context-free.
E. True
In every possible pumping decomposition, pumping produces a string outside $B$.
This contradicts the CFL pumping lemma.
Hence, $B$ is not CFL.
