The major product obtained in the following reaction is:

NaOH, $$\Delta$$

Here an internal Aldol reaction will take place where in a carbanion will be formed at the carbon on the left of ketone rather than right because the formation of a 5 membered ring is more preferrable compared to the formation of a 7 membered ring due to steric factors. This carbanion will in turn attack on the aldehyde group and reduce it to alcohol. Upon heating -OH group will leave with a hydrogen and a double bond will be formed.