If the line frac{x - 1}{2} = frac{y + alpha}{ | vedclass

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(D) Since the line lies in the plane,the point $(1, -\alpha, \beta)$ on the line must satisfy the plane equation $2x + y + z = 5$.Substituting the poi

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$-9$

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(D) Since the line lies in the plane,the point $(1, -\alpha, \beta)$ on the line must satisfy the plane equation $2x + y + z = 5$.Substituting the point into the plane equation: $2(1) + (-\alpha) + \beta = 5 \Rightarrow 2 - \alpha + \beta = 5 \Rightarrow \beta - \alpha = 3$.Also,the direction vector of the line $(2, \alpha, 2)$ must be perpendicular to the normal vector of the plane $(2, 1, 1)$.Thus,the dot product is zero: $2(2) + \alpha(1) + 2(1) = 0$.$4 + \alpha + 2 = 0 \Rightarrow \alpha = -6$.Substituting $\alpha = -6$ into $\beta - \alpha = 3$:$\beta - (-6) = 3 \Rightarrow \beta + 6 = 3 \Rightarrow \beta = -3$.Therefore,$\alpha + \beta = -6 + (-3) = -9$.

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

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