If a→=i^+6j^+3k^;b→=3i^+2j^+k^ and c→=(α+1)i^+(β−1)j^+k^ are linearly dependent vectors and |c→|=6; then the possible value (s) of (α+β) can be:

Take a¯ b¯ c¯=0 ⇒β=3 ∴c→=(α+1)i^+2j^+k^ |c→|=6 (α+1)2+4+1=6 ⇒α=0,−2 ∴α+β=3,1

If a→=i^+6j^+3k^;b→=3i^+2j^+k^ and c→=(α+1)i^+(β−1)j^+k^ are linearly dependent vectors and |c→|=6; then the possible value (s) of (α+β) can be:

Take a¯ b¯ c¯=0 ⇒β=3 ∴c→=(α+1)i^+2j^+k^ |c→|=6 (α+1)2+4+1=6 ⇒α=0,−2 ∴α+β=3,1

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$If \vec{a} = \hat{i} + 6 \hat{j} + 3 \hat{k}; \vec{b} = 3 \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = (α + 1) \hat{i} + (β - 1) \hat{j} + \hat{k} are linearly dependent vectors and | \vec{c} | = \sqrt{6}; then the possible$
$value (s) of (α + β) can be:$

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$Take [\begin{array}{lll} \bar{a} \bar{b} \bar{c} \end{array}] = 0 \Rightarrow β = 3 ∴ \vec{c} = (α + 1) \hat{i} + 2 \hat{j} + \hat{k} | \vec{c} | = \sqrt{6} (α + 1)^{2} + 4 + 1 = 6 \Rightarrow α = 0, - 2 ∴ α + β = 3, 1$

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If a→=i^+6j^+3k^;b→=3i^+2j^+k^ and c→=(α+1)i^+(β−1)j^+k^ are linearly dependent vectors and |c→|=6; then the possible value (s) of (α+β) can be:

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$If \vec{a} = \hat{i} + 6 \hat{j} + 3 \hat{k}; \vec{b} = 3 \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = (α + 1) \hat{i} + (β - 1) \hat{j} + \hat{k} are linearly dependent vectors and | \vec{c} | = \sqrt{6}; then the possible$
$value (s) of (α + β) can be:$