Read the given information carefully and answer the questions below:

1%2 means 2 is neither smaller nor greater than 1
1&2 means 2 is neither greater nor equal to 1
1*2 means 2 is neither smaller nor equal to 1
1$2 means 2 is not greater than 1
1α2 means 2 is not smaller than 1

Statements: P*D$C%B$A; H$T&Y%D; K*UαW%P
Conclusions: I. D&U II. B*T III. H&K

The correct option is All are true.

Let us first decode the symbols given in the problem statement:

1. 1%2 means 2 is neither smaller nor greater than 1 ⇒ 2 = 1 (or 1 = 2).
2. 1&2 means 2 is neither greater nor equal to 1 ⇒ 2 < 1 (or 1 > 2).
3. 1*2 means 2 is neither smaller nor equal to 1 ⇒ 2 > 1 (or 1 < 2).
4. 1$2 means 2 is not greater than 1 ⇒ 2 ≤ 1 (or 1 ≥ 2).
5. 1α2 means 2 is not smaller than 1 ⇒ 2 ≥ 1 (or 1 ≤ 2).

Now, let us convert the given statements into standard mathematical inequality expressions:

Statement 1: P*D$C%B$A
• P*D ⇒ P < D
• D$C ⇒ D ≥ C
• C%B ⇒ C = B
• B$A ⇒ B ≥ A
Combined Statement 1:

$P<D\ge C=B\ge A$

Statement 2: H$T&Y%D
• H$T ⇒ H ≥ T
• T&Y ⇒ T > Y
• Y%D ⇒ Y = D
Combined Statement 2:

$H\ge T>Y=D$

Statement 3: K*UαW%P
• K*U ⇒ K < U
• UαW ⇒ U ≤ W
• W%P ⇒ W = P
Combined Statement 3:

$K<U\le W=P$

Merging all three statements together through common elements (P, D, Y, W):

$K<U\le W=P<D=Y<T\le H$

Also, from D ≥ C = B ≥ A, we know:

$T>Y=D\ge C=B$

Now, let us evaluate each conclusion step-by-step:

Conclusion I: D&U
Decoded form: D > U (i.e., U < D).
From our combined chain:

$U\le W=P<D$

This clearly yields $U<D$ or $D>U$. Thus, Conclusion I is TRUE.

Conclusion II: B*T
Decoded form: B < T (i.e., T > B).
From our combined chain:

$B=C\le D=Y<T$

This clearly yields $B<T$. Thus, Conclusion II is TRUE.

Conclusion III: H&K
Decoded form: H > K (i.e., K < H).
From our combined chain:

$K<U\le W=P<D=Y<T\le H$

This clearly yields $K<H$ or $H>K$. Thus, Conclusion III is TRUE.

Since all conclusions I, II, and III are true, the correct answer option is All are true.