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Question : 4 of 100
Marks:
+1,
-0
Solution:
‌M(3λ+1,2λ+2,−2λ−1) ∴α+β+γ=3λ+2 ‌N(−3µ−2,−2µ+2,4µ+1) ∴a+b+c=−µ+1
‌‌=‌=‌‌3λµ+2µ=3λµ+λ‌2µ=λ‌2λµ−λ=λµ+2µ‌λµ=λ+2µ⇒
‌λµ=2λ ⇒µ=2‌‌(λ≠0)∴‌‌λ=4‌α+β+γ=14‌a+b+c=−1‌‌=196
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