1. Write the given lines in point-direction form

The lines are $L_1: \vec r = (1,2,3) + \lambda(1,-1,1)$ and $L_2: \vec r = (4,5,6) + \mu(1,1,-1).$

So,

• a point on $L_1$ is $A=(1,2,3)$ with direction vector $\vec d_1=(1,-1,1)$,
• a point on $L_2$ is $B=(4,5,6)$ with direction vector $\vec d_2=(1,1,-1)$.

Let the required points be $P=(1+\lambda,\ 2-\lambda,\ 3+\lambda)$ on $L_1$, and $Q=(4+\mu,\ 5+\mu,\ 6-\mu)$ on $L_2$.

2. Use the fact for shortest distance between two skew lines

For the common perpendicular $PQ$, the vector $\overrightarrow{PQ}$ must be perpendicular to both direction vectors $\vec d_1$ and $\vec d_2$.

Now, $\overrightarrow{PQ}=Q-P=(3+\mu-\lambda,\ 3+\mu+\lambda,\ 3-\mu-\lambda).$

So we require: $\overrightarrow{PQ}\cdot \vec d_1=0$ and $\overrightarrow{PQ}\cdot \vec d_2=0.$

3. Form the equations

First, $\overrightarrow{PQ}\cdot \vec d_1=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(-1)+(3-\mu-\lambda)(1)=0.$

Simplifying, $3+\mu-\lambda-3-\mu-\lambda+3-\mu-\lambda=0$ $3-\mu-3\lambda=0.$ So, $\mu+3\lambda=3. \qquad (1)$

Next, $\overrightarrow{PQ}\cdot \vec d_2=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(1)+(3-\mu-\lambda)(-1)=0.$

Simplifying, $3+\mu-\lambda+3+\mu+\lambda-3+\mu+\lambda=0$ $3+3\mu+\lambda=0.$ So, $\lambda+3\mu=-3. \qquad (2)$

4. Solve for $\lambda$ and $\mu$

From (1): $\mu=3-3\lambda.$ Substitute into (2): $\lambda+3(3-3\lambda)=-3$ $\lambda+9-9\lambda=-3$ $-8\lambda=-12$ $\lambda=\frac{3}{2}.$

Then $\mu=3-3\cdot \frac{3}{2}=3-\frac{9}{2}=-\frac{3}{2}.$

5. Find points $P$ and $Q$

For $P$: $P=\left(1+\frac{3}{2},\ 2-\frac{3}{2},\ 3+\frac{3}{2}\right)=\left(\frac{5}{2},\ \frac{1}{2},\ \frac{9}{2}\right).$

For $Q$: $Q=\left(4-\frac{3}{2},\ 5-\frac{3}{2},\ 6+\frac{3}{2}\right)=\left(\frac{5}{2},\ \frac{7}{2},\ \frac{15}{2}\right).$

6. Find the midpoint of $PQ$

If midpoint is $(\alpha,\beta,\gamma)$, then $\alpha=\frac{\frac{5}{2}+\frac{5}{2}}{2}=\frac{5}{2},$ $\beta=\frac{\frac{1}{2}+\frac{7}{2}}{2}=2,$ $\gamma=\frac{\frac{9}{2}+\frac{15}{2}}{2}=6.$

Thus, $(\alpha,\beta,\gamma)=\left(\frac{5}{2},2,6\right).$

7. Compute the required value

$\alpha+\beta+\gamma=\frac{5}{2}+2+6=\frac{5}{2}+8=\frac{21}{2}.$ Therefore, $2(\alpha+\beta+\gamma)=2\cdot \frac{21}{2}=21.$

8. Comparison with stored answer

Derived answer is $21$, which matches the stored correct answer.