Let the point (-1, alpha, beta) lie on the li | vedclass

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(D) Let the two lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = \lambda$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0} = \mu$.Any p

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$25$

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(D) Let the two lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = \lambda$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0} = \mu$.Any point on $L_1$ is $P(-3\lambda-2, 4\lambda+2, 2\lambda+5)$ and any point on $L_2$ is $Q(-\mu-2, 2\mu-6, 1)$.The direction ratios of the line of shortest distance $PQ$ are proportional to the cross product of the direction vectors of $L_1$ and $L_2$,which are $\vec{v_1} = -3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{v_2} = -1\hat{i} + 2\hat{j} + 0\hat{k}$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 4 & 2 \ -1 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0 - (-2)) + \hat{k}(-6 - (-4)) = -4\hat{i} - 2\hat{j} - 2\hat{k}$.This is parallel to the vector $2\hat{i} + \hat{j} + \hat{k}$.The vector $\vec{PQ} = ((-3\lambda-2) - (-\mu-2))\hat{i} + ((4\lambda+2) - (2\mu-6))\hat{j} + ((2\lambda+5) - 1)\hat{k} = (\mu-3\lambda)\hat{i} + (4\lambda-2\mu+8)\hat{j} + (2\lambda+4)\hat{k}$.Since $\vec{PQ}$ is parallel to $2\hat{i} + \hat{j} + \hat{k}$,we have $\frac{\mu-3\lambda}{2} = \frac{4\lambda-2\mu+8}{1} = \frac{2\lambda+4}{1}$.From $\frac{4\lambda-2\mu+8}{1} = \frac{2\lambda+4}{1}$,we get $2\lambda - 2\mu + 4 = 0 \Rightarrow \mu = \lambda + 2$.Substituting $\mu = \lambda + 2$ into $\frac{\mu-3\lambda}{2} = 2\lambda+4$,we get $\frac{\lambda+2-3\lambda}{2} = 2\lambda+4 \Rightarrow -\lambda+1 = 2\lambda+4 \Rightarrow 3\lambda = -3 \Rightarrow \lambda = -1$.Then $\mu = -1+2 = 1$.The line of shortest distance passes through $P(-3(-1)-2, 4(-1)+2, 2(-1)+5) = P(1, -2, 3)$ and $Q(-1-2, 2(1)-6, 1) = Q(-3, -4, 1)$.The equation of the line $PQ$ is $\frac{x-1}{-3-1} = \frac{y-(-2)}{-4-(-2)} = \frac{z-3}{1-3} \Rightarrow \frac{x-1}{-4} = \frac{y+2}{-2} = \frac{z-3}{-2} \Rightarrow \frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{1}$.Since $(-1, \alpha, \beta)$ lies on this line,$\frac{-1-1}{2} = \frac{\alpha+2}{1} = \frac{\beta-3}{1} \Rightarrow -1 = \alpha+2 = \beta-3$.Thus,$\alpha = -3$ and $\beta = 2$.Therefore,$(\alpha-\beta)^2 = (-3-2)^2 = (-5)^2 = 25$.

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

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