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a, b, c with unknowns fixed by two dot products: find the scalar triple product

The same question, asked 2 times: 2021 Paper 7*, 2021 Paper 17. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 7

    85
    If a→=αi^+βj^+3k^\overrightarrow{a}= \alpha\widehat{i}+ \beta\widehat{j}+ 3\widehat{k},
    b→=−βi^−αj^−k^ and c→=i^−2j^−k^\begin{matrix} & \overrightarrow{b}= - \beta\widehat{i}- \alpha\widehat{j}-\widehat{k}\text{~and~} \\ & \overrightarrow{c}=\widehat{i}- 2\widehat{j}-\widehat{k} \end{matrix}
    such that a→⋅b→=1\overrightarrow{a}\cdot\overrightarrow{b}= 1 and b→⋅c→=−3\overrightarrow{b}\cdot\overrightarrow{c}= - 3, then 13((a→×b→)⋅c→)\frac{1}{3}((\overrightarrow{a}\times\overrightarrow{b}) \cdot\overrightarrow{c}) is equal to

  2. Question 2

    2021 Paper 17

    81
    Let a→=i^−αj^+βk^\overrightarrow{a}=\widehat{i}-\alpha\widehat{j}+\beta\widehat{k}, b→=3i^+βj^−αk^\overrightarrow{b}=3\widehat{i}+\beta\widehat{j}-\alpha\widehat{k} and c→=−αi^−2j^+k^\overrightarrow{c}=-\alpha\widehat{i}-2\widehat{j}+\widehat{k}, where α\alpha and β\beta are integers. If a→⋅b→=−1\overrightarrow{a}\cdot\overrightarrow{b}=-1 and b→⋅c→=10\overrightarrow{b}\cdot\overrightarrow{c}=10, then (a→×b→)⋅c→(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c} is equal to: