If sin⁡xsin⁡ysin⁡zcos⁡xcos⁡ycos⁡zcos3⁡xcos3⁡ycos3⁡z=0, then which of the following is/are possible?

sin⁡xsin⁡ysin⁡zcos⁡xcos⁡ycos⁡zcos3⁡xcos3⁡ycos3⁡zTaking cos x, cos y and cos z common from C1, C2 and C3, respectivelyΔ=cos⁡xcos⁡ycos⁡ztan⁡xtan⁡ytan⁡z111cos2⁡xcos2⁡ycos2⁡zApplying C1→C1−C2,C2→C2−C3, we getΔ=cos⁡xcos⁡ycos⁡ztan⁡x−tan⁡ytan⁡y−tan⁡ztan⁡z001cos2⁡x−cos2⁡ycos2⁡y−cos2⁡zcos2⁡z=−cos⁡xcos⁡ycos⁡ztan⁡x−tan⁡ytan⁡y−tan⁡zcos2⁡x−cos2⁡ycos2⁡y−cos2⁡z=−cos⁡xcos⁡ycos⁡z×sin⁡(x−y)cos⁡xcos⁡ysin⁡(y−z)cos⁡ycos⁡z−sin⁡(x−y)sin⁡(x+y)−sin⁡(y−z)sin⁡(y+z)=sin⁡(x−y)sin⁡(y−z)cos⁡zcos⁡xsin⁡(x+y)sin⁡(y+z)=sin⁡(x−y)sin⁡(y−z)[cos⁡zsin⁡(y+z)−cos⁡xsin⁡(x+y)]=sin⁡(x−y)sin⁡(y−z)sin⁡ycos2⁡z+sin⁡zcos⁡ycos⁡z−sin⁡xcos⁡ycos⁡x−sin⁡ycos2⁡x=sin⁡(x−y)sin⁡(y−z)sin⁡ycos2⁡z−cos2⁡x+cos⁡y(sin⁡zcos⁡z−sin⁡xcos⁡x)]=sin⁡(x−y)sin⁡(y−z)[sin⁡ysin⁡(x+z)sin⁡(x−z)+cos⁡y(sin⁡2z−sin⁡2x)/2]=sin⁡(x−y)sin⁡(y−z)[sin⁡ysin⁡(x+z)sin⁡(x−z)+cos⁡ysin⁡(z−x)cos⁡(z+x)]=sin⁡(x−y)sin⁡(y−z)sin⁡(z−x)[−sin⁡ysin⁡(x+z)+cos⁡ycos⁡(z+x)]=sin⁡(x−y)sin⁡(y−z)sin⁡(z−x)cos⁡(x+y+z)For Δ=0,x=y  or y=z  or z=x  or x+y+z=π/2.