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c = xi + (x - 2)j - k lies in the plane of i - j + 2k and i + j + k: find x

The same question, asked 2 times: 15 May 2023 S2, 11 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    15 May 2023 S2

    139
    If the vector c⃗\vec{c} lies in the plane of a⃗\vec{a} and b⃗\vec{b}, where a⃗=i^−j^+2k^, b⃗=i^+j^+k^\vec{a}=\hat{i}-\hat{j}+2\hat{k},\, \vec{b}=\hat{i}+\hat{j}+\hat{k} and c⃗=xi^−(2−x)j^−k^\vec{c}=x\hat{i}-(2-x)\hat{j}-\hat{k}, then the value of xx is

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  2. Question 2

    11 Apr 2026 S2

    87
    If aˉ=i^+j^+k^,bˉ=i^−j^+2k^\bar{a} = \hat{i} + \hat{j} + \hat{k}, \bar{b} = \hat{i} - \hat{j} + 2\hat{k} and cˉ=xi^+(x−2)j^−k^\bar{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k} are three vectors in which cˉ\bar{c} lies in the plane of aˉ\bar{a} and bˉ\bar{b}, then x=⋯x = \cdots

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