Suppose, the line x - 2 = y - 2 - 5 = z + 2 2 lies on the plane x + 3y - 2z + = 0. Then ( + ) is equal to _______.

Given equation of line x - 2 = y - 2 - 5 = z + 2 2 ...... (i) and plane x + 3y - 2z + = 0 ...... (ii) Line (i) passes through (2, 2, -2) which lies on plane (ii). 2 + 6 + 4 + = 0 = - 12 Also, given line is perpendicular to normal of the plane (1) - 5(3) + 2(-2) = 0 = 19 + = 19 + (-12) =

Suppose the line x 2 alpha y 2 5 z 2 2 lies on the plane x 3: null (Mathematics)

Given equation of line

{{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2} \quad ...... (i)

and plane $x + 3y - 2z + \beta = 0 \quad ...... (ii)$

Line (i) passes through $(2, 2, -2)$

which lies on plane (ii).

\therefore 2 + 6 + 4 + \beta = 0

\Rightarrow \beta = - 12

Also, given line is perpendicular to normal of the plane

\alpha (1) - 5(3) + 2(-2) = 0

\Rightarrow \alpha = 19

\therefore \alpha + \beta = 19 + (-12) = 19 - 12 = 7

Explanation

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