See the answer below.
Work Step by Step
a)$\pi$ bonds must occur each in its own plane (overlap of p orbitals). Supposing the three carbons are along the x-axis, one $\pi$ bond occurs parallel to the xz-plane and the other, parallel to the xy-plane. b) The central carbon is $sp$ hybridized because it makes 2 $\pi$ bonds, the others are $sp^2$ because they only make one. c) p orbitals for the $\pi$ bonds, $sp-sp^2$ orbitals in the $\sigma$ bonds.