A line passing through the points P(1, -1, 2) | vedclass

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(D) The equation of the line passing through $P(1, -1, 2)$ and $Q(2, 0, 1)$ is given by $\frac{x-1}{2-1} = \frac{y-(-1)}{0-(-1)} = \frac{z-2}{1-2} = k

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$3\sqrt{3}$

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(D) The equation of the line passing through $P(1, -1, 2)$ and $Q(2, 0, 1)$ is given by $\frac{x-1}{2-1} = \frac{y-(-1)}{0-(-1)} = \frac{z-2}{1-2} = k$, which simplifies to $\frac{x-1}{1} = \frac{y+1}{1} = \frac{z-2}{-1} = k$.Thus, any point on the line is $(k+1, k-1, -k+2)$.For point $A$ on the $XY$-plane, the $z$-coordinate is $0$: $-k+2 = 0 \implies k = 2$. So, $A = (2+1, 2-1, 0) = (3, 1, 0)$.For point $B$ on the $YZ$-plane, the $x$-coordinate is $0$: $k+1 = 0 \implies k = -1$. So, $B = (-1+1, -1-1, -(-1)+2) = (0, -2, 3)$.The distance $AB = \sqrt{(3-0)^2 + (1-(-2))^2 + (0-3)^2} = \sqrt{3^2 + 3^2 + (-3)^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$.

$A$ line passing through the points $P(1, -1, 2)$ and $Q(2, 0, 1)$ meets the $XY$-plane and the $YZ$-plane at points $A$ and $B$ respectively. The distance $AB$ is equal to...

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