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 If x, y, z are different from zero and Δ=ab-yc-za-xbc-za-xb-yc=0 then the value of the expression  ax+by+cz is 

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detailed solution

Correct option is D

Applying , R¯2→R¯2-R¯1,R¯3→R¯3-R¯1 we get Δ=ab-yc-z-xy0-x0z=0Expanding along C3, we get(c-z)-xy-x0+zab-y-xy=0⇒  (c-z)(xy)+z(ay+bx-xy)=0⇒  cxy-xyz+ayz+bxz-xyz=0⇒  ayz+bzx+cxy=2xyz⇒  ax+by+cz=2


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