The line L passes through the point (1, 2, 3) | vedclass

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(D) The distance between two lines is constant if and only if the lines are parallel. Given line $L_1: \vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$. The

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$(-5, 6, 2)$

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(D) The distance between two lines is constant if and only if the lines are parallel. Given line $L_1: \vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$. The direction vector of $L_1$ is $\vec{v} = (3, -2, 1)$. Since line $L$ is parallel to $L_1$,its direction vector must also be $\vec{v} = (3, -2, 1)$. Given that $L$ passes through $(1, 2, 3)$,the equation of line $L$ is $\vec{r} = (1, 2, 3) + t(3, -2, 1)$. This can be written in parametric form as $x = 1 + 3t, y = 2 - 2t, z = 3 + t$. We check which point does not satisfy this equation: For $(4, 0, 4)$: $4 = 1 + 3t \implies t = 1$. Then $y = 2 - 2(1) = 0$ and $z = 3 + 1 = 4$. This point lies on $L$. For $(-2, 4, 2)$: $-2 = 1 + 3t \implies t = -1$. Then $y = 2 - 2(-1) = 4$ and $z = 3 - 1 = 2$. This point lies on $L$. For $(7, -2, 5)$: $7 = 1 + 3t \implies t = 2$. Then $y = 2 - 2(2) = -2$ and $z = 3 + 2 = 5$. This point lies on $L$. For $(-5, 6, 2)$: $-5 = 1 + 3t \implies t = -2$. Then $y = 2 - 2(-2) = 6$ and $z = 3 - 2 = 1$. Since $z = 1 
eq 2$,this point does not lie on $L$.

The line $L$ passes through the point $(1, 2, 3)$. The distance of any point on the line $L$ from the line $\vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$ is constant. Then the line $L$ does not pass through the point:

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