Correct Answer – A
माना `a=lambdai+(1+lambda)hatj+(1+2lambda)hatk`
तथा `b=(1-lambda)hati+._(1)lambdahatj+2hatk`
अब, a तथा b के परस्पर लम्ब होने के लिए, `a.b=|a||b|cos””(pi)/(2)=0`
`Rightarrow [lambdahati+(1+lambda)hatj+(1+2lambda)hatk]`
`[(1-lambda)hati+lambdaj+2hatk]=0`
`Rightarrow lambda(1-lambda)+(1+lambda)lambda+(1+2lambda)2=0`
`Rightarrow lambda-lambda^(2)+lambda+lambda^(2)+2+4lambda=0`
`Rightarrow 2+6lambda=0 Rightarrow 6lambda=0`
`Rightarrow lambda=-(2)/(6)=(1)/(3)`
`therefore lambda=-(1)/(3)`