Hybridization of central carbon is $\text{sp}$ and two of its unhybridized p -orbitals are involved in $\pi$ - bonding. Also p - orbitals at an atom are orthogonal (at ${90}^{\text{ο}}$ angle), the two $\pi$ - bonds at central carbon are also orthogonal. Let these p - orbitals, involved in $\pi$ - bonding, at central carbon be ${\text{p}}_{\text{y}}$ and ${\text{p}}_{\text{z}}$, then

 

Hybridization at the terminal carbons are both ${\text{sp}}^2$ . The two p - orbitals involved in ${\text{sp}}^2$ hybridization are ${\text{p}}_{\text{x}}$ and ${\text{p}}_{\text{y}}$ at left terminal and ${\text{p}}_{\text{x}}$ and ${\text{p}}_{\text{z}}$ at right terminals. Therefore, the three $\text{s}{\text{p}}^2$ hybrid orbitals at left terminal carbon are in $\text{x}\text{y}$ plane, while the same at right terminal carbon are in $\text{x}\text{z}$
plane. As a result, the two triangular planes at terminals are at right angles as shown below: 

      
The two $\text{H - C - H}$ planes at terminals are perpendicular to each other. ($\pi$ - bonds are also perpendicular to each other.)
      Hence, (a) Molecule is non - planar. (b) It doesn't show geometrical isomerism because two H and chlorine are in different planes. (c) polar.