A line with direction ratios $$1, -1, 2$$ intersects the lines $$\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$$ and $$\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$$ at the points $$P$$ and $$Q$$, respectively. If the length of the line segment $$PQ$$ is $$\alpha$$, then $$225\alpha^2$$ is equal to :

Let the required line $$L$$ have direction ratios $$1,-1,2$$.
Take a general point $$R(a,b,c)$$ on this line; then every point on $$L$$ is

$$\mathbf{r}= (a,b,c) + \lambda(1,-1,2) \qquad (\lambda\in\mathbb{R})$$

1. Coordinates of intersection point $$P$$ with the line $$\dfrac{x}{2}=\dfrac{y}{3}= \dfrac{z+1}{3}=s_1$$ are

$$P(2s_1,\;3s_1,\;3s_1-1).$$

Since $$P$$ also lies on $$L$$, there is a parameter $$\lambda$$ such that

$$\bigl(2s_1,\,3s_1,\,3s_1-1\bigr)= (a,b,c)+\lambda(1,-1,2).$$

2. Coordinates of intersection point $$Q$$ with the line $$\dfrac{x+1}{-1}=\dfrac{y-2}{1}= \dfrac{z}{4}=s_2$$ are

$$Q(-s_2-1,\;s_2+2,\;4s_2).$$

There is a parameter $$\mu$$ such that

$$\bigl(-s_2-1,\,s_2+2,\,4s_2\bigr)= (a,b,c)+\mu(1,-1,2).$$

3. Subtract the two position vectors to eliminate $$(a,b,c)$$:

$$Q-P=(\mu-\lambda)(1,-1,2).$$

Writing this component-wise gives three equations

$$\begin{aligned} -s_2-1-2s_1 &= \mu-\lambda \quad -(1)\\ \;s_2+2-3s_1 &= -(\mu-\lambda) \quad -(2)\\ \;4s_2-3s_1+1 &= 2(\mu-\lambda) \quad -(3) \end{aligned}$$

Add $$(1)$$ and $$(2)$$ to eliminate $$\mu-\lambda$$:

$$(-s_2-1-2s_1)+(s_2+2-3s_1)=0 \;\Longrightarrow\; 1-5s_1=0 \;\Longrightarrow\; s_1=\dfrac15.$$

From $$(1)$$,

$$\mu-\lambda = -s_2-1-2\!\left(\dfrac15\right)= -s_2-\dfrac{7}{5}. \quad -(4)$$

Insert $$s_1=1/5$$ in $$(3)$$:

$$4s_2-\dfrac35+1 = 2(\mu-\lambda) \;\Longrightarrow\; 4s_2+\dfrac25 = 2(\mu-\lambda). \quad -(5)$$

Substitute $$(4)$$ into $$(5)$$:

$$4s_2+\dfrac25 = 2\!\left(-s_2-\dfrac75\right) \;\Longrightarrow\; 4s_2+\dfrac25 = -2s_2-\dfrac{14}{5} \;\Longrightarrow\; 6s_2 = -\dfrac{16}{5} \;\Longrightarrow\; s_2 = -\dfrac{8}{15}.$$

Now $$\mu-\lambda = -\!\left(-\dfrac{8}{15}\right)-\dfrac75= \dfrac{8}{15}-\dfrac{21}{15}= -\dfrac{13}{15}.$"

4. Compute coordinates of $$P$$ and $$Q$$:

$$P\!$$\left$$(2\!$$\left$$(\dfrac15$$\right$$),\,3\!$$\left$$(\dfrac15$$\right$$),\,3\!$$\left$$(\dfrac15$$\right$$)-1$$\right$$)=$$\left$$(\dfrac25,\dfrac35,-\dfrac25$$\right$$),$$

$$Q\!$$\left$$(-\!$$\left$$(-\dfrac{8}{15}$$\right$$)-1,\;-\dfrac{8}{15}+2,\;4\!$$\left$$(-\dfrac{8}{15}$$\right$$)$$\right$$)=$$\left$$(-\dfrac{7}{15},\dfrac{22}{15},-\dfrac{32}{15}$$\right$$).$$

5. Vector $$\overrightarrow{PQ}=Q-P$$ is

$$$$\left$$(-\dfrac{7}{15}-\dfrac25,\;\dfrac{22}{15}-\dfrac35,\;-\dfrac{32}{15}+\dfrac25$$\right$$)=$$\left$$(-\dfrac{13}{15},\;\dfrac{13}{15},\;-\dfrac{26}{15}$$\right$$).$$

6. Length $$$$\alpha$$=\lVert\overrightarrow{PQ}\rVert$$:

$$$$\alpha^2 =\left$$(\dfrac{-13}{15}$$\right$$)^2+$$\left$$(\dfrac{13}{15}$$\right$$)^2+$$\left$$(\dfrac{-26}{15}$$\right$$)^2 = \dfrac{169+169+676}{225} = \dfrac{1014}{225}.$$

Therefore $$225$$\alpha^2$$ = 1014.$$

Hence the correct choice is:
Option B which is: $$1014$$