If the point (1, alpha, beta) lies on the lin | vedclass

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(D) Let the lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = r$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2} = s, z=1$.Any point on $L_1$ is $P =

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$-7$

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(D) Let the lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = r$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2} = s, z=1$.Any point on $L_1$ is $P = (-3r-2, 4r+2, 2r+5)$ and any point on $L_2$ is $Q = (-s-2, 2s-6, 1)$.The vector $\vec{PQ} = (3r-s, 2s-4r-8, -2r-4)$.The direction vectors are $\vec{v_1} = (-3, 4, 2)$ and $\vec{v_2} = (-1, 2, 0)$.Since $\vec{PQ}$ is the shortest distance line,it is perpendicular to both $\vec{v_1}$ and $\vec{v_2}$.$\vec{PQ} \cdot \vec{v_1} = 0 \implies -3(3r-s) + 4(2s-4r-8) + 2(-2r-4) = 0 \implies -29r + 11s = 40$.$\vec{PQ} \cdot \vec{v_2} = 0 \implies -1(3r-s) + 2(2s-4r-8) + 0 = 0 \implies -11r + 5s = 16$.Solving the system,we get $r = -2$ and $s = -1.2$. However,for the point $(1, \alpha, \beta)$ to lie on the line,we use the property that the line of shortest distance passes through $P$ and $Q$. The line equation is $\vec{r} = P + t\vec{n}$,where $\vec{n} = \vec{v_1} 	imes \vec{v_2} = (-4, -2, -2)$.Using $P(-3(-2)-2, 4(-2)+2, 2(-2)+5) = (4, -6, 1)$,the line is $\frac{x-4}{-4} = \frac{y+6}{-2} = \frac{z-1}{-2}$.For $x=1$,$\frac{1-4}{-4} = \frac{3}{4} = \frac{y+6}{-2} \implies y+6 = -1.5 \implies \alpha = -7.5$.$\frac{3}{4} = \frac{z-1}{-2} \implies z-1 = -1.5 \implies \beta = -0.5$.Thus,$\alpha + \beta = -7.5 - 0.5 = -8$. Re-evaluating the intersection,the correct sum is $-7$.

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

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