Correct Answer - D
दिया है `(axxb)+(bxxc)+(cxxa)=lambda(bxxc)`
`Rightarrow (axxb)+(1-lambda)(bxxc)+(cxxa)=0`
`Rightarrow -(2b+3c)xxb+(1-lambda)(bxxc)+(cxxa)=0[therefore a+2b+3c=0]`
`Rightarrow 3(bxxc)+(1-lambda)(bxxc)+cxx(-2b-3c)=0`
`[therefore bxxb=0 bxxc=-cxxb]`
`Rightarrow (3+1-lambda)(bxxc)-2(cxxb)-3(cxxc)=0`
`Rightarrow 2(bxxc)+(4-lambda)(bxxc)-3(cxxc)=0`
`Rightarrow (4-lambda+2)(bxxc)=0`
`Rightarrow (6-lambda)(bxxc)=0[therefore bxxc ne 0]`
`Rightarrow (6-lambda)=0 Rightarrow lambda=6`
