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(C) To determine if all atoms lie in one plane,we analyze the hybridization and geometry of each molecule:$(A)$ $CH_2=CH-CH=CH_2$ ($1,3$-butadiene): D

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$(B,C)$

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(C) To determine if all atoms lie in one plane,we analyze the hybridization and geometry of each molecule:$(A)$ $CH_2=CH-CH=CH_2$ ($1,3$-butadiene): Due to rotation around the central $C-C$ single bond,it can exist in $s$-cis and $s$-trans conformations. While the $s$-cis form is planar,the molecule can rotate,and not all conformations are necessarily planar.$(B)$ $HC\equiv C-CH=CH_2$ (but$-1-$en$-3-$yne): The $HC\equiv C-$ group is linear ($sp$-hybridized). The $-CH=CH_2$ group is planar ($sp^2$-hybridized). Since the linear group is attached to the planar group,the entire molecule remains planar.$(C)$ $H_2C=C=O$ (ketene): The central carbon is $sp$-hybridized,but the terminal $CH_2$ group is $sp^2$-hybridized. The two $H$-atoms on the $CH_2$ group lie in the same plane as the $C=C=O$ backbone.$(D)$ $H_2C=C=CH_2$ (allene): The central carbon is $sp$-hybridized. The two terminal $CH_2$ groups lie in perpendicular planes to each other,making the molecule non-planar.Thus,in compounds $(B)$ and $(C)$,all atoms lie in the same plane.

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):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

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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)$ $H_2C=C=CH_2$