Given :
{(→a×→b)×(→b×→c)}×(→c×→a)=→0⇒{[→a →b →c]→b−[→a →b →b]→c}×(→c×→a)=→0⇒[→a →b →c]((→a⋅→b)→c−(→b⋅→c)→a)=→0∵[→a →b →b]=0⇒[→a →b →c]=0 OR →a⋅→b=→b⋅→c=0⇒∣∣
∣∣α2−312α−22−α1∣∣
∣∣=0
OR α+4α+6=2−2α2−2=0⇒10−15α=0
OR α=0=−65 (contradictory,not possible)
⇒α=1015=23