What is the number of $sp^3$ hybridized carbon atoms in $HO(CH_2)_2CH(CH_3)_2$?

Amongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

In the compound $CH \equiv C - CH = CH_2$,what is the hybridization of the carbons involved in the $C - C$ single bond?

The hybridisation of the negatively charged carbon atoms in $(1)$ and $(2)$ are:
$(1)$ $CH_3^-$
$(2)$ $H_2C^- - CHO - CH_3$

Consider a positively charged $C_2H_3$ species in which the positively charged carbon is $sp$-hybridized,the uncharged carbon is $sp^2$-hybridized and an empty $p$-orbital is perpendicular to the $\pi$ system. What is the best description of this cation?

In which of the compounds given below is there more than one kind of hybridisation $(sp, sp^2, sp^3)$ for carbon?
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$