A line L passes through points A(1, 3, 2) and | vedclass

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(D) The direction vector of line $L$ is $\vec{v} = B - A = (2-1, 2-3, 1-2) = (1, -1, -1)$.The equation of line $L$ is $\vec{r} = (1, 3, 2) + t(1, -1,

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$\frac{23}{3}$

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(D) The direction vector of line $L$ is $\vec{v} = B - A = (2-1, 2-3, 1-2) = (1, -1, -1)$.The equation of line $L$ is $\vec{r} = (1, 3, 2) + t(1, -1, -1) = (1+t, 3-t, 2-t)$.Let $M$ be the projection of point $P(1, 1, -1)$ on line $L$. $M$ corresponds to some parameter $t$ on the line,so $M = (1+t, 3-t, 2-t)$.The vector $\vec{PM} = M - P = (1+t-1, 3-t-1, 2-t-(-1)) = (t, 2-t, 3-t)$.Since $\vec{PM}$ is perpendicular to the line $L$,$\vec{PM} \cdot \vec{v} = 0$.$(t)(1) + (2-t)(-1) + (3-t)(-1) = 0 \implies t - 2 + t - 3 + t = 0 \implies 3t = 5 \implies t = \frac{5}{3}$.The coordinates of $M$ are $(1+\frac{5}{3}, 3-\frac{5}{3}, 2-\frac{5}{3}) = (\frac{8}{3}, \frac{4}{3}, \frac{1}{3})$.Let the mirror image of $P$ be $P'(x, y, z)$. Since $M$ is the midpoint of $PP'$,we have $M = \frac{P+P'}{2}$,so $P' = 2M - P$.$x = 2(\frac{8}{3}) - 1 = \frac{16}{3} - \frac{3}{3} = \frac{13}{3}$.$y = 2(\frac{4}{3}) - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3}$.$z = 2(\frac{1}{3}) - (-1) = \frac{2}{3} + \frac{3}{3} = \frac{5}{3}$.Thus,$x+y+z = \frac{13}{3} + \frac{5}{3} + \frac{5}{3} = \frac{23}{3}$.

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$

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