Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} $ and $ \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} $. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

<p>To find the distance between the points $ P $ and $ Q $, let's consider the following steps:</p> <p><p><strong>Determine the direction vector of line $ L $:</strong> </p> <p>For the line $ L $ to be perpendicular to both given lines, we find the direction vector of $ L $ using the cross product of the direction vectors of the given lines.</p> <p>The first line has direction vector $ \langle 2, 1, -2 \rangle $ and the second line has direction vector $ \langle 1, 3, 4 \rangle $. The cross product is:</p> <p>$ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k} = 5(2\hat{i} - 2\hat{j} + \hat{k}) $</p> <p>Thus, the direction vector of line $ L $ is $ \langle 2, -2, 1 \rangle $.</p></p> <p><p><strong>Equation of line $ L $:</strong> </p> <p>Since $ L $ passes through point $ P(2, -1, 3) $ with direction vector $ \langle 2, -2, 1 \rangle $, its parametric form is:</p> <p>$ \frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda $</p></p> <p><p><strong>Find the intersection with the $ yz $-plane:</strong> </p> <p>On the $ yz $-plane, $ x = 0 $. Substitute into the line equation to find $ \lambda $:</p> <p>$ 2\lambda + 2 = 0 \implies \lambda = -1 $</p> <p>Substitute $ \lambda = -1 $ back into the parametric equations to find $ Q $:</p> <p>$ x = 2(-1) + 2 = 0, \quad y = -2(-1) - 1 = 1, \quad z = (-1) + 3 = 2 $</p> <p>Therefore, the intersection point $ Q $ is $ (0, 1, 2) $.</p></p> <p><p><strong>Calculate the distance between $ P $ and $ Q $:</strong></p> <p>$ \text{d}(P, Q) = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $</p></p> <p>Thus, the distance between points $ P $ and $ Q $ is 3.</p>