Step 1: Identify the structures E, F, and G.

The problem asks for the relationships between three structures E, F, and G. Based on the options provided (tautomers, geometrical isomers, diastereomers), this is a typical isomerism question. The structures from the original JEE paper are:

• Structure E: $CH_3-CO-CH_2-CHO$ (3-oxobutanal). This is a dicarbonyl compound, specifically a $\beta$-ketoaldehyde.

• Structure F: (Z)-3-hydroxybut-2-enal. This is an enol form of structure E. The structure is depicted as:

  $$\begin{array}{ccc} \text{CH}_3 & & \text{CHO} \\ & \diagdown & \diagup \\ & \text{C} = & \text{C} \\ & \diagup & \diagdown \\ \text{OH} & & \text{H} \end{array}$$

  This is the Z-isomer because the higher priority groups on each carbon of the double bond (-OH and -CHO) are on the same side.

• Structure G: (E)-3-hydroxybut-2-enal. This is also an enol form of structure E, and a stereoisomer of F. The structure is depicted as:

  $$\begin{array}{ccc} \text{CH}_3 & & \text{H} \\ & \diagdown & \diagup \\ & \text{C} = & \text{C} \\ & \diagup & \diagdown \\ \text{OH} & & \text{CHO} \end{array}$$

  This is the E-isomer because the higher priority groups (-OH and -CHO) are on opposite sides.

Step 2: Evaluate Option A - Resonance Structures.

Resonance structures are different representations of the electron distribution in a single molecule. The positions of the atoms must be the same in all resonance structures.

• The conversion of E to F (or G) involves the movement of a hydrogen atom from the central carbon to the oxygen atom, along with a shift in the position of a double bond. Since atoms have moved, E, F, and G are not resonance structures. They are isomers.
• F and G have the same connectivity but different spatial arrangements of atoms, so they are stereoisomers, not resonance structures.
• Therefore, statement A is incorrect.

Step 3: Evaluate Option B - Tautomers.

Tautomers are constitutional isomers that readily interconvert, typically through the migration of a proton. Keto-enol tautomerism is a common example.

• Structure E is the keto form ($CH_3-CO-CH_2-CHO$).
• Structures F and G are enol forms ($CH_3-C(OH)=CH-CHO$).
• The interconversion between E and F (or E and G) is a keto-enol tautomerization. For example, to get from E to F: $CH_3-CO-CH_2-CHO \rightleftharpoons CH_3-C(OH)=CH-CHO$
• This involves the migration of a proton from the $\alpha$-carbon (the $CH_2$ group) to the keto oxygen.
• Since both F and G are enol forms of the keto structure E, E and F are tautomers, and E and G are also tautomers.
• Therefore, statement B is correct.

Step 4: Evaluate Option C - Geometrical Isomers.

Geometrical isomers (cis-trans or E/Z isomers) are stereoisomers that differ in the spatial arrangement of groups around a double bond or a ring structure.

• Structure F is (Z)-3-hydroxybut-2-enal.
• Structure G is (E)-3-hydroxybut-2-enal.
• They have the same molecular formula ($C_4H_6O_2$) and the same connectivity of atoms. However, they differ in the arrangement of the groups (-$CH_3$, -OH, -H, -CHO) around the C=C double bond.
• Therefore, F and G are geometrical isomers.
• Statement C is correct.

Step 5: Evaluate Option D - Diastereomers.

Diastereomers are stereoisomers that are not mirror images of each other.

• F and G are stereoisomers.
• To check if they are enantiomers (mirror images), let's draw the mirror image of F. The mirror image of the Z-isomer is still a Z-isomer. F and G are Z and E isomers, respectively. They are not mirror images of each other.
• A general rule is that geometrical isomers are a type of diastereomer.
• Since F and G are geometrical isomers, they are also diastereomers.
• Therefore, statement D is correct.

Conclusion:

Based on the analysis of each option:

• A is incorrect.
• B is correct.
• C is correct.
• D is correct.

The correct statements are B, C, and D.