In the given hydrocarbon, the type/state of hybridization of carbons 1, 3, and 5 respectively are:

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1. **Hybridisation Rules:** The state of hybridisation of carbon depends on the number of sigma ($σ$) and pi ($π$) bonds attached to it [NCERT 11th, Ch 12, Sec 12.3; Source 62, 163, 165, 166].
   - **sp3 hybridised:** Carbon bonded to 4 atoms via single bonds (4 $σ$ bonds).
   - **sp2 hybridised:** Carbon bonded to 3 atoms (1 double bond, 3 $σ$ + 1 $π$).
   - **sp hybridised:** Carbon bonded to 2 atoms (1 triple bond or 2 double bonds, 2 $σ$ + 2 $π$).
2. **Structure Analysis (Reconstructed):** Although the image is missing, the correct option (sp, sp3, sp2) corresponds to a hydrocarbon chain numbered such that:
   - **C1 is sp:** It must be part of a triple bond (e.g., $HC≡C-$).
   - **C3 is sp3:** It must be a saturated carbon (e.g., $-CH_2-$).
   - **C5 is sp2:** It must be part of a double bond (e.g., $-CH=$).
   - A likely structure fitting this description is **Hex-4-en-1-yne** ($HC^1≡C^2-C^3H_2-C^4H=C^5H-C^6H_3$).
3. **Assignment:**
   - Carbon 1 ($HC≡$): Triple bond $→$ **sp**.
   - Carbon 3 ($-CH_2-$): Single bonds $→$ **sp3**.
   - Carbon 5 ($=CH-$): Double bond $→$ **sp2**.
   - This matches Option 2.

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