Dear Aspirant
(2i^+6j^+27k^)×(i^+λj^+μk^)=0
( i^ j^ k^)
( 2 6 27 ) =0i^+0j^+0k^
( 1 λ μ )
i^(6μ−27λ)−j^(2μ−27)+k^(2λ−6)=0i^+0j^+0k^
On comparing the corresponding components, we have:
6μ−27λ=0
2μ−27=0
2λ−6=0
Now,
2λ−6=0⇒λ=3
2μ−27=0⇒μ=27/2
λ=3 and μ=27/2 ans
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