In case of three plane mirrors meeting at a point to form a corner of a cube, if incident light suffers one reflection on each mirror;

Explanation of Ans (1)

Let incident ray =$a\hat{\mathcal{i}}+b\hat{\mathcal{j}}+c\hat{\mathcal{k}}$

Reflection take place on $\mathcal{x}-\mathcal{y}$ plane then reflected ray will be 

= $a\hat{\mathcal{i}}+b\hat{\mathcal{j}}-c\hat{\mathcal{k}}$ after reflection only $\mathcal{z}$ component will get reverse.

$\therefore$     Reflected ray = $a\hat{\mathcal{i}}+b\hat{\mathcal{j}}-c\hat{\mathcal{k}}$

          Now on second reflection $\mathcal{y}-\mathcal{z}$ plane incident ray is $a\hat{\mathcal{i}}+b\hat{\mathcal{j}}-c\hat{\mathcal{k}}$ this reflected ray is $-a\hat{\mathcal{i}}+b\hat{\mathcal{j}}-c\hat{\mathcal{k}}$.

          Third reflection will give reflected ray=$-a\hat{\mathcal{i}}-b\hat{\mathcal{j}}-c\hat{\mathcal{k}}$

$\therefore$      Angle between final reflected ray & incident ray is $180°$.

          Explanation of Ans(4)

          Due to reflection on plane phase of light change by $\pi$.

$\therefore$      Emergent ray will be out of phase with incident ray.