Distance Between Intersection Points of a Line with Two Lines

To find the length $\alpha$ of the line segment $PQ$, we start by parameterizing points $P$ and $Q$ on the given lines.

Let $P$ be a point on line $L_1: \frac{x}{2} = \frac{y}{3} = \frac{z+1}{3} = \lambda$. The general coordinates of $P$ can be written as: $P = (2\lambda, 3\lambda, 3\lambda - 1)$

Let $Q$ be a point on line $L_2: \frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4} = \mu$. The general coordinates of $Q$ can be written as: $Q = (-\mu - 1, \mu + 2, 4\mu)$

The vector $\vec{PQ}$ joining point $P$ to point $Q$ is given by: $\vec{PQ} = Q - P = (-\mu - 1 - 2\lambda)\hat{i} + (\mu + 2 - 3\lambda)\hat{j} + (4\mu - 3\lambda + 1)\hat{k}$

Since the line passing through $P$ and $Q$ has direction ratios $1, -1, 2$, the vector $\vec{PQ}$ must be parallel to $\hat{i} - \hat{j} + 2\hat{k}$. Thus, its components are proportional to these direction ratios: $\frac{-\mu - 1 - 2\lambda}{1} = \frac{\mu + 2 - 3\lambda}{-1} = \frac{4\mu - 3\lambda + 1}{2}$

Equating the first two ratios: $\frac{-\mu - 1 - 2\lambda}{1} = \frac{\mu + 2 - 3\lambda}{-1}$ $-1(-\mu - 1 - 2\lambda) = 1(\mu + 2 - 3\lambda)$ $\mu + 1 + 2\lambda = \mu + 2 - 3\lambda$ $5\lambda = 1 \implies \lambda = \frac{1}{5}$

Equating the first and third ratios: $\frac{-\mu - 1 - 2\lambda}{1} = \frac{4\mu - 3\lambda + 1}{2}$ $2(-\mu - 1 - 2\lambda) = 4\mu - 3\lambda + 1$ $-2\mu - 2 - 4\lambda = 4\mu - 3\lambda + 1$ $6\mu = -\lambda - 3$

Substituting $\lambda = \frac{1}{5}$ into this equation: $6\mu = -\frac{1}{5} - 3 = -\frac{16}{5} \implies \mu = -\frac{8}{15}$

Now, let $\vec{PQ} = k(\hat{i} - \hat{j} + 2\hat{k})$. The scalar $k$ is equal to the first component of $\vec{PQ}$: $k = -\mu - 1 - 2\lambda = -\left(-\frac{8}{15}\right) - 1 - 2\left(\frac{1}{5}\right) = \frac{8}{15} - 1 - \frac{2}{5} = -\frac{13}{15}$

The length of the line segment $PQ$ is given by $\alpha = |\vec{PQ}|$: $\alpha = |k| \sqrt{1^2 + (-1)^2 + 2^2} = \left|-\frac{13}{15}\right| \sqrt{6} = \frac{13\sqrt{6}}{15}$

Squaring both sides: $\alpha^2 = \frac{169 \times 6}{225}$

Thus, the value of $225\alpha^2$ is: $225\alpha^2 = 225 \times \frac{169 \times 6}{225} = 169 \times 6 = 1014$