Let A=a−2b+3c,B=−4a+5b−6c and C=xa−9b+zc AB‌=‌ Position vector of ‌B−‌ Position vector of ‌A ‌=−4a+5b−6c−a+2b−3c ‌=−5a+7b−9c BC‌=‌ Position vector of ‌C−‌ Position vector of ‌B ‌=xa−9b+zc+4a−5b+6c ‌=(x+4)a−14b+(z+6)c A,B and Care collinear if AB=λBC ‌−5a+7b−9c=λ((x+4)a−14b+(z+6)c) ∴‌7=−14λ⇒λ=−‌
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⇒−5=(x+4)λ and −9=λ(z+6) ‌⇒‌‌−5=(x+4)(−‌
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) and 9=−‌
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(z+6) ‌⇒‌‌10=x+4 and 18=z+6⇒x=6 and z=12 Hence, 2x−z=2×6−12=0