The correct option is
B a=b′×c′[a′b′c′]Given
→a⋅a′=→b⋅b′=→c⋅c′=1
→a⋅b′=→a⋅c′=→b⋅a′=→b⋅c′=→c⋅a′=→c⋅b′=0
a+b+c=1a′+1b′+1c′
by taking LCM
a+b+c=b′×c′+c′×a′+a′×b′(a′×b′)⋅(a′×c′)
a+b+c=b′×c′+c′×a′+a′×b′a′⋅(b′×c′)
a+b+c=b′×c′+c′×a′+a′×b′[a′b′c′]
a+b+c=b′×c′[a′b′c′]+c′×a′[a′b′c′]+a′×b′[a′b′c′]
comparing both sides
→a=b′×c′[a′b′c′]
b=c′×a′[a′b′c′]
c=a′×b′[a′b′c′]