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If the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$ then $(α, β)$ =

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6,−17

-6,7

5,−16

−5,15

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Correct option is B

The condition that the line x−x1l=y−y1m=z−z1nto lies on the plane ax+by+cz+d=0 is al+bm+cn=0and ax1+by1+cz1+d=0Hence, 13+3−5−α2=03−15−2α=02α=−12α=−6And 12+31+6−2+β=05−12+β=0β=7 Therefore, α,β=−6,7

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If the line x−23=y−1−5=z+22 lies in the plane x+3y−αz+β=0 then α,β=

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If the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$ then $(α, β)$ =