Key Concepts and Formulas

1. Equation of a Line in Vector Form: A line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ can be written as $\vec{r} = \vec{a} + t\vec{b}$, where $t$ is a scalar parameter. From the Cartesian form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$, we have $\vec{a} = (x_1, y_1, z_1)$ and $\vec{b} = (l, m, n)$.
2. Shortest Distance Between Two Skew Lines: The shortest distance $D$ between two skew lines $\vec{r}_1 = \vec{a}_1 + t\vec{b}_1$ and $\vec{r}_2 = \vec{a}_2 + s\vec{b}_2$ is given by the formula: $D = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{||\vec{b}_1 \times \vec{b}_2||}$ This formula represents the projection of the vector connecting any two points on the lines onto the common perpendicular direction of the lines.
3. Parametric Form of a Line: A point $(x,y,z)$ on a line given by $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ can be expressed in terms of a parameter, say $k$, as $x = x_1 + lk$, $y = y_1 + mk$, $z = z_1 + nk$.

Step-by-Step Solution

Step 1: Convert the given lines into vector form. We are given two lines in Cartesian form. To use the shortest distance formula, we first convert them to vector form $\vec{r} = \vec{a} + t\vec{b}$.

For line $L$: $\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}$ This line passes through the point $(5, \lambda, -\lambda)$ and is parallel to the direction vector $(-2, 0, 1)$. So, $\vec{a}_1 = (5, \lambda, -\lambda)$ and $\vec{b}_1 = (-2, 0, 1)$.

For line $L_1$: $x+1=y-1=4-z$ We need to rewrite this in the standard symmetric form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$. $\frac{x-(-1)}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$ This line passes through the point $(-1, 1, 4)$ and is parallel to the direction vector $(1, 1, -1)$. So, $\vec{a}_2 = (-1, 1, 4)$ and $\vec{b}_2 = (1, 1, -1)$.

Step 2: Calculate the vector connecting the points on the lines, $\vec{a}_2 - \vec{a}_1$. This vector connects a point on $L_1$ to a point on $L$. $\vec{a}_2 - \vec{a}_1 = (-1-5, 1-\lambda, 4-(-\lambda)) = (-6, 1-\lambda, 4+\lambda)$

Step 3: Calculate the cross product of the direction vectors, $\vec{b}_1 \times \vec{b}_2$. This cross product gives a vector that is perpendicular to both lines, representing the direction of the shortest distance. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix}$ $= \mathbf{i}((0)(-1) - (1)(1)) - \mathbf{j}((-2)(-1) - (1)(1)) + \mathbf{k}((-2)(1) - (0)(1))$ $= \mathbf{i}(0-1) - \mathbf{j}(2-1) + \mathbf{k}(-2-0)$ $= -\mathbf{i} - \mathbf{j} - 2\mathbf{k} = (-1, -1, -2)$

Step 4: Calculate the magnitude of the cross product, $||\vec{b}_1 \times \vec{b}_2||$. This is the magnitude of the common perpendicular direction vector, which forms the denominator of the shortest distance formula. $||\vec{b}_1 \times \vec{b}_2|| = \sqrt{(-1)^2 + (-1)^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$

Step 5: Calculate the scalar triple product component, $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$. This is the dot product of the vector connecting the points on the lines with the common perpendicular vector. Its absolute value forms the numerator of the shortest distance formula. $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-6, 1-\lambda, 4+\lambda) \cdot (-1, -1, -2)$ $= (-6)(-1) + (1-\lambda)(-1) + (4+\lambda)(-2)$ $= 6 - 1 + \lambda - 8 - 2\lambda$ $= -3 - \lambda$

Step 6: Use the shortest distance formula to solve for $\lambda$. We are given that the shortest distance $D = 2\sqrt{6}$. $D = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{||\vec{b}_1 \times \vec{b}_2||}$ $2\sqrt{6} = \frac{|-3 - \lambda|}{\sqrt{6}}$ Multiply both sides by $\sqrt{6}$: $2\sqrt{6} \cdot \sqrt{6} = |-3 - \lambda|$ $2 \cdot 6 = |-(3 + \lambda)|$ $12 = |3 + \lambda|$ This gives two possible cases: Case 1: $3 + \lambda = 12 \implies \lambda = 9$ Case 2: $3 + \lambda = -12 \implies \lambda = -15$ Since the problem states $\lambda \geq 0$, we must choose $\lambda = 9$.

Step 7: Establish the relationship between $\alpha$ and $\gamma$ for a point $(\alpha, \beta, \gamma)$ on line L. A general point on line $L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}$ can be expressed parametrically. Let this common ratio be $k$. $\frac{x-5}{-2} = k \implies x = 5-2k$ $\frac{y-\lambda}{0} = k \implies y-\lambda = 0 \implies y = \lambda$ $\frac{z+\lambda}{1} = k \implies z = k-\lambda$ Since $(\alpha, \beta, \gamma)$ lies on $L$ and we found $\lambda = 9$: $\alpha = 5-2k$ $\beta = 9$ $\gamma = k-9$ From the equation for $\gamma$, we can express $k$ in terms of $\gamma$: $k = \gamma+9$. Substitute this expression for $k$ into the equation for $\alpha$: $\alpha = 5-2(\gamma+9)$ $\alpha = 5-2\gamma-18$ $\alpha = -13-2\gamma$ Rearranging this, we get the fundamental relationship between $\alpha$ and $\gamma$: $\alpha + 2\gamma = -13$

Step 8: Check which of the given options is NOT possible using the derived relationship. We will substitute the relationship $\alpha = -13-2\gamma$ into each option or compare the option directly with $\alpha + 2\gamma = -13$.

(A) $\alpha+2 \gamma=24$ Comparing this with our derived relationship $\alpha+2\gamma=-13$, we see that $24 \neq -13$. Therefore, this option is NOT possible.

(B) $2 \alpha+\gamma=7$ Substitute $\alpha = -13-2\gamma$: $2(-13-2\gamma) + \gamma = -26 - 4\gamma + \gamma = -26 - 3\gamma$. If $-26 - 3\gamma = 7$, then $-3\gamma = 33 \implies \gamma = -11$. This is a possible value for $\gamma$, so this option is possible.

(C) $\alpha-2 \gamma=19$ Substitute $\alpha = -13-2\gamma$: $(-13-2\gamma) - 2\gamma = -13 - 4\gamma$. If $-13 - 4\gamma = 19$, then $-4\gamma = 32 \implies \gamma = -8$. This is a possible value for $\gamma$, so this option is possible.

(D) $2 \alpha-\gamma=9$ Substitute $\alpha = -13-2\gamma$: $2(-13-2\gamma) - \gamma = -26 - 4\gamma - \gamma = -26 - 5\gamma$. If $-26 - 5\gamma = 9$, then $-5\gamma = 35 \implies \gamma = -7$. This is a possible value for $\gamma$, so this option is possible.

Thus, the only option that is not possible is (A).

Common Mistakes & Tips

• Incorrectly converting line equations: Pay close attention to the signs and coefficients when converting from Cartesian to vector form (e.g., $4-z$ becomes $\frac{z-4}{-1}$, not $\frac{z-4}{1}$). Also, lines like $\frac{y-\lambda}{0}$ imply $y-\lambda=0$, so $y=\lambda$.
• Algebraic errors in vector calculations: Double-check dot products, cross products, and magnitudes to avoid calculation mistakes.
• Forgetting the absolute value: The shortest distance must always be positive, so remember to take the absolute value of the scalar triple product in the numerator.
• Ignoring constraints: The condition $\lambda \geq 0$ is crucial for determining the correct value of $\lambda$.
• Parametric representation of a point: Ensure you correctly express the coordinates of a point on the line in terms of a parameter, especially when one of the direction ratios is zero.

Summary

The problem required us to find a constraint on the coordinates of a point lying on a given line, using the information about the shortest distance to another line. We first converted both lines into their vector forms to identify their position and direction vectors. Then, we applied the shortest distance formula between two skew lines, which involved calculating the vector connecting points on the lines, the cross product of their direction vectors, and their dot product. Solving the shortest distance equation yielded the value of the parameter $\lambda$. Finally, we used this value of $\lambda$ to find the relationship between the coordinates $(\alpha, \beta, \gamma)$ of a point on the first line. By comparing this derived relationship with the given options, we identified the option that was not possible.

The final answer is $\boxed{A}$