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a = 2i + 2j + 3k, b = -i + 2j + k, c = 3i + j: lambda making a + lambda b perpendicular to c

The same question, asked 3 times: 2 May 2023 S1, 10 May 2023 S2*, 16 May 2023 S2. Every wording below, oldest first.

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  1. Question 1

    2 May 2023 S1

    124
    If a⃗=(2i^+2j^+3k^)\vec{a} = (2\hat{i} + 2\hat{j} + 3\hat{k}), b⃗=(−i^+2j^+k^)\vec{b} = (-\hat{i} + 2\hat{j} + \hat{k}) and c⃗=(3i^+j^)\vec{c} = (3\hat{i} + \hat{j}) such that (a⃗+λb⃗)(\vec{a} + \lambda\vec{b}) is perpendicular to c⃗\vec{c}, then the value of λ\lambda is

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  2. Question 2

    10 May 2023 S2

    141
    If a⃗=2i^+2j^+3k^\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k}, b⃗=−i^+2j^+k^\vec{b} = -\hat{i}+2\hat{j}+\hat{k} and c⃗=3i^+j^\vec{c} = 3\hat{i}+\hat{j} such that b⃗+λa⃗\vec{b}+\lambda\vec{a} is perpendicular to c⃗\vec{c}, then λ\lambda is

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  3. Question 3

    16 May 2023 S2

    131
    If a⃗=2i^+2j^+3k^\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k}, b⃗=−i^+2j^+k^\vec{b} = -\hat{i}+2\hat{j}+\hat{k} and c⃗=3i^+j^\vec{c} = 3\hat{i}+\hat{j} are vectors such that a⃗+λb⃗\vec{a}+\lambda\vec{b} is perpendicular to c⃗\vec{c}, then value of λ\lambda is

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