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Question

If the line x-23=y-1-5=z+22 lies in the plane x+3y-αz+7=0, then α = _________________.

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Solution

Given line x-23=y-1-5=z+22
Lies in the plane x + 3y – αz + 7 = 0
Whose normal n=i^+3j^-αk^ then b=3i^-5j^+2k^n=i^+3j^-αk^i.e. 31-53+2-α=0i.e. 3-15-2α=0i.e. -12-2α=0i.e. 2α=-12i.e. α=-6

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