Coplanarity requires the scalar triple product to be zero. Computing the determinant:
1(−2λ+5)+3(−4λ+3)+2(10−3)=0
−2λ+5−12λ+9+14=0
−14λ+28=0, so λ=2.
Let the vectors a=i^−3j^+2k^,b=2i^+j^−k^ and c=3i^+5j^−2λk^ be coplanar. Then λ is equal to
Held on 25 May 2023 · Verified 13 Jul 2026.
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