If the line frac{x-2}{3}=frac{y-1}{-5}=frac{z | vedclass

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(B) The given equation of the line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$.This line passes through the point $P(2, 1, -2)$.Since the line lie

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$(-6,7)$

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(B) The given equation of the line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$.This line passes through the point $P(2, 1, -2)$.Since the line lies in the plane $x+3y-\alpha z+\beta=0$,the point $P$ must satisfy the plane equation:$2 + 3(1) - \alpha(-2) + \beta = 0$$2 + 3 + 2\alpha + \beta = 0$$2\alpha + \beta = -5$  $(i)$Also,the direction vector of the line $\vec{v} = (3, -5, 2)$ is perpendicular to the normal vector of the plane $\vec{n} = (1, 3, -\alpha)$.Therefore,their dot product is zero:$3(1) + (-5)(3) + 2(-\alpha) = 0$$3 - 15 - 2\alpha = 0$$-12 - 2\alpha = 0$$2\alpha = -12 \implies \alpha = -6$Substituting $\alpha = -6$ into equation $(i)$:$2(-6) + \beta = -5$$-12 + \beta = -5$$\beta = 7$Thus,$(\alpha, \beta) = (-6, 7)$.

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

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