### Related Formula The vector connecting the shortest distance points P$P$ and Q$Q$ on two skew lines must be simultaneously perpendicular to the direction vectors vecb_1$\vec{b}_1$ and vecb_2$\vec{b}_2$ of both lines: vecPQ parallel (vecb_1 times vecb_2)$\vec{PQ} \parallel (\vec{b}_1 \times \vec{b}_2)$ ### Core Logic Let's define general points on both lines: - Point P$P$ on L_1$L_1$: (1+lambda, \, 2-lambda, \, 3+lambda)$(1+\lambda, \, 2-\lambda, \, 3+\lambda)$ - Point Q$Q$ on L_2$L_2$: (4+mu, \, 5+mu, \, 6-mu)$(4+\mu, \, 5+\mu, \, 6-\mu)$ The direction ratios of vector vecPQ$\vec{PQ}$ are: vecPQ = (3+mu-lambda)hati + (3+mu+lambda)hatj + (3-mu-lambda)hatk$\vec{PQ} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}$ ### Step 1: Compute Perpendicular Direction Vector Calculate the cross product of the directions of lines L_1$L_1$ and L_2$L_2$: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 1 & -1 & 1 \\ 1 & 1 & -1 endvmatrix = 0hati + 2hatj + 2hatk$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 0\hat{i} + 2\hat{j} + 2\hat{k}$ Since vecPQ$\vec{PQ}$ is parallel to (0, 2, 2)$(0, 2, 2)$, we compare the coordinate ratios: 3+mu-lambda = 0 implies lambda - mu = 3 quad implies (1)$3+\mu-\lambda = 0 \implies \lambda - \mu = 3 \quad \implies (1)$ frac3+mu+lambda2 = frac3-mu-lambda2 implies 2mu + 2lambda = 0 implies lambda + mu = 0 quad implies (2)$\frac{3+\mu+\lambda}{2} = \frac{3-\mu-\lambda}{2} \implies 2\mu + 2\lambda = 0 \implies \lambda + \mu = 0 \quad \implies (2)$

### Step 2: Solve for Parameters and Midpoint Solving linear equations (1) and (2) simultaneously: lambda = frac32, quad mu = -frac32$\lambda = \frac{3}{2}, \quad \mu = -\frac{3}{2}$ Substitute these values back to find the specific coordinates of points P$P$ and Q$Q$: - P = left(frac52, \, frac12, \, frac92 ight)$P = \left(\frac{5}{2}, \, \frac{1}{2}, \, \frac{9}{2} ight)$ - Q = left(frac52, \, frac72, \, frac152 ight)$Q = \left(\frac{5}{2}, \, \frac{7}{2}, \, \frac{15}{2} ight)$ The midpoint coordinates (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ are: (alpha, beta, gamma) = left( frac5/2 + 5/22, \, frac1/2 + 7/22, \, frac9/2 + 15/22 right) = left(frac52, \, 2, \, 6 ight)$(\alpha, \beta, \gamma) = \left( \frac{5/2 + 5/2}{2}, \, \frac{1/2 + 7/2}{2}, \, \frac{9/2 + 15/2}{2} \right) = \left(\frac{5}{2}, \, 2, \, 6 ight)$ ### Step 3: Final Computation Calculate the required terms: 2(alpha+beta+gamma) = 2left(frac52 + 2 + 6right) = 5 + 4 + 12 = 21$2(\alpha+\beta+\gamma) = 2\left(\frac{5}{2} + 2 + 6\right) = 5 + 4 + 12 = 21$ ### Pattern Recognition Sees: Explicit endpoints of the shortest distance line vector segment. Shortcut: Since the cross product component along hati$\hat{i}$ is 0, the x-coordinates of both line points are identical, providing a massive shortcut to check algebraic equations immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry