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(C) The point $(1, \mu, 2)$ lies on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$.Setting $x=1$,we get $\frac{1-4}{1} = -3$,so $\frac{\mu-9

Vedclass · Accepted Answer

$\frac{\sqrt{146}}{7}$

Vedclass · Answer

(C) The point $(1, \mu, 2)$ lies on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$.Setting $x=1$,we get $\frac{1-4}{1} = -3$,so $\frac{\mu-9}{2} = -3 \Rightarrow \mu = 3$ and $\frac{2-5}{1} = -3$. Thus,the foot is $(1, 3, 2)$.The vector from $(\lambda, 2, 3)$ to $(1, 3, 2)$ is $(1-\lambda, 3-2, 2-3) = (1-\lambda, 1, -1)$.Since this vector is perpendicular to the line direction $(1, 2, 1)$,we have $(1-\lambda)(1) + (1)(2) + (-1)(1) = 0$,which gives $1-\lambda + 1 = 0 \Rightarrow \lambda = 2$.The lines are $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $L_2: \frac{x-2}{2} = \frac{y-3}{3} = \frac{z+5}{6}$.These are parallel lines with direction vector $\vec{v} = (2, 3, 6)$.The distance between parallel lines is $d = \frac{|(\vec{r_2}-\vec{r_1}) 	imes \vec{v}|}{|\vec{v}|}$.Here,$\vec{r_1} = (1, 2, -4)$ and $\vec{r_2} = (2, 3, -5)$,so $\vec{r_2}-\vec{r_1} = (1, 1, -1)$.The cross product is $(1, 1, -1) 	imes (2, 3, 6) = (6 - (-3), -(6 - (-2)), 3-2) = (9, -8, 1)$.The magnitude is $\sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81+64+1} = \sqrt{146}$.The magnitude of $\vec{v}$ is $\sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.Thus,the distance is $\frac{\sqrt{146}}{7}$.

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