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Let the vectors a⃗=i^−3j^+2k^,b⃗=2i^+j^−k^\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}, \vec{b} = 2\hat{i} + \hat{j} - \hat{k} and c⃗=3i^+5j^−2λk^\vec{c} = 3\hat{i} + 5\hat{j} - 2\lambda\hat{k} be coplanar. Then λ\lambda is equal to

Solution

✅ Correct Option: 4

Coplanarity requires the scalar triple product to be zero. Computing the determinant:

1(−2λ+5)+3(−4λ+3)+2(10−3)=01(-2\lambda + 5) + 3(-4\lambda + 3) + 2(10 - 3) = 0

−2λ+5−12λ+9+14=0-2\lambda + 5 - 12\lambda + 9 + 14 = 0

−14λ+28=0-14\lambda + 28 = 0, so λ=2\lambda = 2.

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