What is 3\alpha+2\beta equal to if (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\alpha\hat{j}+\beta\hat{k}) is a null vector?

Explanation

For the cross product to be a null vector, the two vectors must be collinear. Thus, their corresponding components are proportional: \frac{2}{1} = \frac{6}{\alpha} = \frac{27}{\beta}. Solving these gives \alpha = 3 and \beta = 13.5. Then 3\alpha + 2\beta = 3(3) + 2(13.5) = 9 + 27 = 36.

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