Let the line of the shortest distance between the lines $$L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$$ and $$L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$$ intersect $$L_1$$ and $$L_2$$ at $$P$$ and $$Q$$ respectively. If $$(\alpha, \beta, \gamma)$$ is the midpoint of the line segment $$PQ$$, then $$2(\alpha + \beta + \gamma)$$ is equal to:

Find $$2(\alpha + \beta + \gamma)$$ where $$(\alpha, \beta, \gamma)$$ is the midpoint of PQ, the shortest distance line between $$L_1$$ and $$L_2$$. The line $$L_1$$ passes through $$A=(1,2,3)$$ with direction vector $$\vec{b_1}=(1,-1,1)$$, and $$L_2$$ passes through $$B=(4,5,6)$$ with direction vector $$\vec{b_2}=(1,1,-1)$$.

Any point on $$L_1$$ can be written as $$P=(1+\lambda,2-\lambda,3+\lambda)$$ and any point on $$L_2$$ as $$Q=(4+\mu,5+\mu,6-\mu)$$, and the vector between them is

$$\vec{PQ}=(3+\mu-\lambda,\;3+\mu+\lambda,\;3-\mu-\lambda)\,.$$

Perpendicularity to $$\vec{b_1}$$ gives

$$\vec{PQ}\cdot\vec{b_1}=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(-1)+(3-\mu-\lambda)(1)=0$$

which simplifies to

$$\mu + 3\lambda = 3 \quad \cdots (1)\,.$$

Similarly, perpendicularity to $$\vec{b_2}$$ yields

$$\vec{PQ}\cdot\vec{b_2}=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(1)+(3-\mu-\lambda)(-1)=0$$

leading to

$$3\mu + \lambda = -3 \quad \cdots (2)\,.$$

From (1) we have $$\mu = 3 - 3\lambda$$, and substituting into (2) gives

$$3(3-3\lambda)+\lambda=-3\implies9-9\lambda+\lambda=-3\implies-8\lambda=-12\implies\lambda=3/2$$

so that $$\mu = 3 - 9/2 = -3/2$$.

Substituting back yields

$$P = (1+3/2,\,2-3/2,\,3+3/2) = (5/2,\,1/2,\,9/2),\quad Q = (4-3/2,\,5-3/2,\,6+3/2) = (5/2,\,7/2,\,15/2)\,.$$

The midpoint $$(\alpha,\beta,\gamma)$$ of PQ is

$$\alpha = \frac{5/2+5/2}{2} = \frac{5}{2},\quad \beta = \frac{1/2+7/2}{2} = 2,\quad \gamma = \frac{9/2+15/2}{2} = 6\,.$$

Therefore,

$$2(\alpha+\beta+\gamma)=2\bigl(\tfrac{5}{2}+2+6\bigr)=2\bigl(\tfrac{5}{2}+8\bigr)=21$$

and the answer is 21.