Key Concepts and Formulas

• Equation of a Line in 3D Space: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(l, m, n)$ can be represented in Cartesian form as $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$. In vector form, it is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector of the line.
• Shortest Distance Between Two Skew Lines: For two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$, the shortest distance $d$ between them is given by the formula: $d = \frac{\left|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\right|}{\left|\vec{b_1} \times \vec{b_2}\right|}$ The numerator represents the magnitude of the scalar triple product of the vectors $(\vec{a_2} - \vec{a_1})$, $\vec{b_1}$, and $\vec{b_2}$. This can also be calculated as the absolute value of the determinant: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$ where $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are points on $L_1$ and $L_2$ respectively, and $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are their respective direction ratios. The denominator is the magnitude of the cross product of the direction vectors, representing the area of the parallelogram formed by $\vec{b_1}$ and $\vec{b_2}$.

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Step-by-Step Solution

Step 1: Extract Position and Direction Vectors from the Given Lines We are given two lines in Cartesian form. We first convert them into vector form to identify the position vectors of points on the lines ($\vec{a_1}, \vec{a_2}$) and their respective direction vectors ($\vec{b_1}, \vec{b_2}$).

The first line $L_1$ is $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$.

• A point on $L_1$ is $A_1(3, \alpha, 3)$. So, its position vector is $\vec{a_1} = 3\hat{i} + \alpha\hat{j} + 3\hat{k}$.
• The direction vector of $L_1$ is $\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$.

The second line $L_2$ is $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$.

• A point on $L_2$ is $A_2(-3, -7, \beta)$. So, its position vector is $\vec{a_2} = -3\hat{i} - 7\hat{j} + \beta\hat{k}$.
• The direction vector of $L_2$ is $\vec{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.

Step 2: Calculate the Vector Connecting Points on the Lines We need the vector $\vec{a_1} - \vec{a_2}$ (or $\vec{a_2} - \vec{a_1}$; the absolute value in the formula handles the sign difference). We will use $\vec{a_1} - \vec{a_2}$ for consistency with the determinant calculation in later steps. $\vec{a_1} - \vec{a_2} = (3\hat{i} + \alpha\hat{j} + 3\hat{k}) - (-3\hat{i} - 7\hat{j} + \beta\hat{k})$ $\vec{a_1} - \vec{a_2} = (3 - (-3))\hat{i} + (\alpha - (-7))\hat{j} + (3 - \beta)\hat{k}$ $\vec{a_1} - \vec{a_2} = 6\hat{i} + (\alpha+7)\hat{j} + (3-\beta)\hat{k}$

Step 3: Determine the Common Perpendicular Direction Vector ($\vec{b_1} \times \vec{b_2}$) The cross product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$ gives a vector that is perpendicular to both lines, representing the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}$ Expanding the determinant: $= \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3))$ $= \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3)$ $= -6\hat{i} - 15\hat{j} + 3\hat{k}$

Step 4: Calculate the Magnitude of the Common Perpendicular Direction Vector ($|\vec{b_1} \times \vec{b_2}|$ ) This magnitude will be the denominator in the shortest distance formula. $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + (3)^2}$ $= \sqrt{36 + 225 + 9} = \sqrt{270}$

Step 5: Calculate the Scalar Triple Product ($(\vec{a_1} - \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2})$) This value forms the numerator (before taking the absolute value) in the shortest distance formula. It can be computed as a determinant using the components of $\vec{a_1} - \vec{a_2}$, $\vec{b_1}$, and $\vec{b_2}$. $(\vec{a_1} - \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2}) = \begin{vmatrix} 6 & \alpha+7 & 3-\beta \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}$ Expanding the determinant along the first row: $= 6((-1)(4) - (1)(2)) - (\alpha+7)((3)(4) - (1)(-3)) + (3-\beta)((3)(2) - (-1)(-3))$ $= 6(-4 - 2) - (\alpha+7)(12 + 3) + (3-\beta)(6 - 3)$ $= 6(-6) - (\alpha+7)(15) + (3-\beta)(3)$ $= -36 - 15\alpha - 105 + 9 - 3\beta$ $= -15\alpha - 3\beta - 132$

Step 6: Apply the Shortest Distance Formula and Solve for $5\alpha + \beta$ The shortest distance $d$ is given as $3\sqrt{30}$. Substitute the calculated values into the formula: $d = \frac{\left|(\vec{a_1} - \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2})\right|}{\left|\vec{b_1} \times \vec{b_2}\right|}$ $3\sqrt{30} = \frac{|-15\alpha - 3\beta - 132|}{\sqrt{270}}$ Now, we solve for the expression involving $\alpha$ and $\beta$: $|-15\alpha - 3\beta - 132| = 3\sqrt{30} \cdot \sqrt{270}$ $|-15\alpha - 3\beta - 132| = 3\sqrt{30 \times 270}$ $|-15\alpha - 3\beta - 132| = 3\sqrt{8100}$ $|-15\alpha - 3\beta - 132| = 3 \times 90$ $|-15\alpha - 3\beta - 132| = 270$ Factor out $-3$ from the expression inside the absolute value: $|-3(5\alpha + \beta + 44)| = 270$ Since $|-3| = 3$: $3|5\alpha + \beta + 44| = 270$ $|5\alpha + \beta + 44| = \frac{270}{3}$ $|5\alpha + \beta + 44| = 90$ This equation leads to two possible cases:

1. $5\alpha + \beta + 44 = 90$ $5\alpha + \beta = 90 - 44$ $5\alpha + \beta = 46$
2. $5\alpha + \beta + 44 = -90$ $5\alpha + \beta = -90 - 44$ $5\alpha + \beta = -134$

The question asks for the positive value of $5\alpha+\beta$. Therefore, $5\alpha+\beta = 46$.

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Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when extracting coordinates from the Cartesian equation of a line, especially for terms like $x+3$ (which corresponds to $x-(-3)$). Also, be meticulous with signs during determinant expansion and vector arithmetic.
• Absolute Value: Always remember to take the absolute value of the scalar triple product in the numerator of the shortest distance formula, as distance must be non-negative. Forgetting this can lead to an incorrect final sign or missing a valid solution.
• Calculation Accuracy: Double-check all calculations, especially those involving cross products, magnitudes, and determinant expansions. Even a small arithmetic error can propagate and lead to an incorrect final answer.

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Summary

This problem required us to find a positive expression involving two unknown parameters, $\alpha$ and $\beta$, given the shortest distance between two skew lines. We began by identifying the position and direction vectors for each line from their Cartesian equations. Then, we calculated the vector connecting points on the lines and the cross product of the direction vectors, which gives the common perpendicular direction. These values were used to compute the scalar triple product and the magnitude of the cross product. Finally, we substituted these into the shortest distance formula, which was provided as $3\sqrt{30}$, and solved the resulting equation for $|5\alpha + \beta + 44|$. Considering the positive value, we found $5\alpha + \beta = 46$.

The final answer is $\boxed{46}$ which corresponds to option (A).