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

The equation of the line passing through $(1, -1, 2)$ and $(2, 0, 1)$ is given by:

$\dfrac{x - 1}{2 - 1} = \dfrac{y - (-1)}{0 - (-1)} = \dfrac{z - 2}{1 - 2}$

$\Rightarrow \dfrac{x - 1}{1} = \dfrac{y + 1}{1} = \dfrac{z - 2}{-1} = \lambda$

Any point on this line can be written as $(\lambda + 1, \lambda - 1, -\lambda + 2)$.

The line meets the XY plane at point A, where the z-coordinate is zero:

$-\lambda + 2 = 0 \Rightarrow \lambda = 2$

Substituting $\lambda = 2$, the coordinates of A are $(3, 1, 0)$.

The line meets the YZ plane at point B, where the x-coordinate is zero:

$\lambda + 1 = 0 \Rightarrow \lambda = -1$

Substituting $\lambda = -1$, the coordinates of B are $(0, -2, 3)$.

The distance between points A and B is:

$AB = \sqrt{(3 - 0)^2 + (1 - (-2))^2 + (0 - 3)^2}$

$AB = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$

Answer: $3\sqrt{3}$