The reaction scheme image/details are not visible in the prompt, so I must infer the intended transformation from the reagent choices.

1. Analyze the three-step reagent patterns

Each option gives three reagents $A$, $B$, and $C$.

Step $A$: $\mathrm{LiAlH_4}$ vs DIBAL-H

• $\mathrm{LiAlH_4}$ is a strong reducing agent.
  • It reduces esters, acids, acyl chlorides, etc. generally all the way to alcohols.
• DIBAL-H is a controlled reducing agent.
  • It reduces esters/nitriles under controlled conditions to aldehydes.

So if the intermediate after step $A$ must be a carbonyl compound (especially aldehyde), DIBAL-H is the more suitable reagent.

Step $B$: $\mathrm{NaOH_{(aq)}}$ vs $\mathrm{NaOH_{(alc)}}$

• $\mathrm{NaOH_{(aq)}}$ generally favors hydrolysis/substitution.
• $\mathrm{NaOH_{(alc)}}$ generally favors elimination to form an alkene.

Thus if the conversion requires formation of a double bond before the final step, alcoholic NaOH is appropriate.

Step $C$: $\mathrm{NH_2NH_2/KOH}$, ethylene glycol vs $\mathrm{Zn/HCl}$

These correspond to the two classic methods for deoxygenation of carbonyl compounds:

• $\mathrm{NH_2NH_2/KOH}$, ethylene glycol: Wolff–Kishner reduction
• $\mathrm{Zn/HCl}$: Clemmensen reduction

Both reduce aldehydes/ketones to methylene groups.

So step $C$ must involve reduction of a carbonyl group to hydrocarbon.

2. Most plausible reaction logic

Among the options, the chemically most coherent sequence for a typical JEE conversion is:

1. DIBAL-H creates an aldehyde selectively.
2. Alcoholic NaOH carries out elimination to generate unsaturation.
3. Zn/HCl reduces the carbonyl to $-CH_2-$.

This matches Option D.

3. Why other options are less suitable

Option A

• $\mathrm{LiAlH_4}$ would over-reduce many carbonyl derivatives to alcohols, not aldehydes.
• $\mathrm{NaOH_{(aq)}}$ is less consistent if elimination is needed.

Option B

• DIBAL-H is fine for selective reduction, but $\mathrm{NaOH_{(aq)}}$ is not suited if alkene formation/elimination is required.

Option C

• $\mathrm{LiAlH_4}$ again is too strong and usually gives alcohols.
• Though alcoholic NaOH and $\mathrm{Zn/HCl}$ are individually plausible, the first step is generally inconsistent.

Option D

• DIBAL-H gives a controlled carbonyl intermediate.
• Alcoholic NaOH promotes elimination.
• Clemmensen reduction removes the carbonyl oxygen.

Hence this is the best fit.

4. Final answer

Therefore, the correct option is:

$\boxed{\text{D}}$

with

$A=\mathrm{DIBAL-H},\quad B=\mathrm{NaOH_{(alc)}},\quad C=\mathrm{Zn/HCl}$