1. Key Concepts and Formulas

• Shortest Distance Between Two Skew Lines: For two lines given by $\vec{r} = \vec{a_1} + t\vec{d_1}$ and $\vec{r} = \vec{a_2} + s\vec{d_2}$, the shortest distance $D$ is given by the formula: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})}{|\vec{d_1} \times \vec{d_2}|} \right|$ Here, $\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on the lines, and $\vec{d_1}$ and $\vec{d_2}$ are their direction vectors.
• Equation and Radius of a Circle: The general equation of a circle in 2D is $x^2 + y^2 + 2gx + 2fy + c = 0$. The center of the circle is $(-g, -f)$ and its radius $R$ is given by $R = \sqrt{g^2 + f^2 - c}$. To find the equation, we substitute the coordinates of three distinct points on the circle into the general equation and solve for $g, f, c$.

2. Step-by-Step Solution

Part 1: Finding the values of $\lambda$

Step 1: Identify points and direction vectors for the given lines. The first line is $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$.

• A point on this line: $\vec{a_1} = (1, 2, 3)$
• Its direction vector: $\vec{d_1} = (2, 3, 4)$

The second line is $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$.

• A point on this line: $\vec{a_2} = (\lambda, 4, 5)$
• Its direction vector: $\vec{d_2} = (3, 4, 5)$

Step 2: Calculate the vector $\vec{a_2} - \vec{a_1}$. This vector connects a point on the first line to a point on the second line, which is essential for calculating the shortest distance. $\vec{a_2} - \vec{a_1} = (\lambda - 1, 4 - 2, 5 - 3) = (\lambda - 1, 2, 2)$

Step 3: Calculate the cross product of the direction vectors, $\vec{d_1} \times \vec{d_2}$. The cross product yields a vector perpendicular to both direction vectors. Its magnitude is the denominator of the shortest distance formula. $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ $= \mathbf{i}(3 \times 5 - 4 \times 4) - \mathbf{j}(2 \times 5 - 4 \times 3) + \mathbf{k}(2 \times 4 - 3 \times 3)$ $= \mathbf{i}(15 - 16) - \mathbf{j}(10 - 12) + \mathbf{k}(8 - 9)$ $= -\mathbf{i} + 2\mathbf{j} - \mathbf{k} = (-1, 2, -1)$

Step 4: Calculate the magnitude of the cross product, $|\vec{d_1} \times \vec{d_2}|$. This is the magnitude of the vector found in Step 3. $|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Step 5: Calculate the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})$. This dot product forms the numerator of the shortest distance formula. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (\lambda - 1)(-1) + (2)(2) + (2)(-1)$ $= -\lambda + 1 + 4 - 2 = -\lambda + 3$

Step 6: Substitute into the shortest distance formula and solve for $\lambda$. We are given that the shortest distance $D = \frac{1}{\sqrt{6}}$. $\frac{1}{\sqrt{6}} = \left| \frac{-\lambda + 3}{\sqrt{6}} \right|$ Multiplying both sides by $\sqrt{6}$: $1 = |-\lambda + 3|$ This absolute value equation gives two possibilities:

• Case 1: $-\lambda + 3 = 1 \implies -\lambda = -2 \implies \lambda = 2$
• Case 2: $-\lambda + 3 = -1 \implies -\lambda = -4 \implies \lambda = 4$ Thus, the values of $\lambda$ are $\lambda_1 = 2$ and $\lambda_2 = 4$.

Part 2: Finding the radius of the circle

Step 7: Identify the three points for the circle. The problem states the circle passes through $(0, 0)$, $(\lambda_1, \lambda_2)$, and $(\lambda_2, \lambda_1)$. Using $\lambda_1 = 2$ and $\lambda_2 = 4$, the three points are:

• $P_1 = (0, 0)$
• $P_2 = (2, 4)$
• $P_3 = (4, 2)$

Step 8: Substitute the points into the general equation of a circle ($x^2 + y^2 + 2gx + 2fy + c = 0$).

• For $P_1(0, 0)$: $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$

• For $P_2(2, 4)$ (with $c=0$): $2^2 + 4^2 + 2g(2) + 2f(4) = 0$ $4 + 16 + 4g + 8f = 0$ $20 + 4g + 8f = 0$ Dividing by 4: $5 + g + 2f = 0 \implies g + 2f = -5$ (Equation 1)

• For $P_3(4, 2)$ (with $c=0$): $4^2 + 2^2 + 2g(4) + 2f(2) = 0$ $16 + 4 + 8g + 4f = 0$ $20 + 8g + 4f = 0$ Dividing by 4: $5 + 2g + f = 0 \implies 2g + f = -5$ (Equation 2)

Step 9: Solve the system of linear equations for $g$ and $f$. We have:

1. $g + 2f = -5$
2. $2g + f = -5$

Subtract Equation 1 from Equation 2 (multiplied by 2): $(4g + 2f) - (g + 2f) = -10 - (-5)$ $3g = -5 \implies g = -\frac{5}{3}$

Substitute $g = -\frac{5}{3}$ into Equation 1: $-\frac{5}{3} + 2f = -5$ $2f = -5 + \frac{5}{3} = -\frac{15}{3} + \frac{5}{3} = -\frac{10}{3}$ $f = -\frac{5}{3}$ So, we have $g = -\frac{5}{3}$, $f = -\frac{5}{3}$, and $c = 0$.

Step 10: Calculate the radius $R$ of the circle. Using the radius formula $R = \sqrt{g^2 + f^2 - c}$: $R = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(-\frac{5}{3}\right)^2 - 0}$ $R = \sqrt{\frac{25}{9} + \frac{25}{9}}$ $R = \sqrt{\frac{50}{9}}$ $R = \frac{\sqrt{50}}{\sqrt{9}} = \frac{5\sqrt{2}}{3}$

3. Common Mistakes & Tips

• Shortest Distance Formula: Ensure correct identification of $\vec{a_1}, \vec{d_1}, \vec{a_2}, \vec{d_2}$ from Cartesian equations. Pay close attention to signs during cross product and dot product calculations. Remember the absolute value for distance.
• Circle Equation: The point $(0,0)$ on the circle always simplifies the problem by directly giving $c=0$.
• Symmetry: For points like $(\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$, the center of the circle will lie on the line $y=x$, implying $g=f$. This can be a useful shortcut to solve for $g$ and $f$ more quickly.
• Collinearity Check: Always check if the three points are collinear. If they are, a unique circle cannot be formed. (In this case, $(0,0), (2,4), (4,2)$ are not collinear since the slope of $P_1P_2$ is $4/2 = 2$ and the slope of $P_1P_3$ is $2/4 = 1/2$).

4. Summary

This problem required a two-part approach. First, we used the formula for the shortest distance between two skew lines to determine the possible values of $\lambda$. This involved careful vector algebra, including cross products and dot products, leading to $\lambda_1 = 2$ and $\lambda_2 = 4$. Second, we used these values to identify three points $(0,0), (2,4), (4,2)$ and then found the radius of the circle passing through them. By substituting these points into the general equation of a circle, we solved for the coefficients $g, f, c$ (where $c=0$ due to the origin being a point on the circle) and subsequently calculated the radius using the formula $R = \sqrt{g^2 + f^2 - c}$. The final calculated radius is $\frac{5\sqrt{2}}{3}$.

5. Final Answer

The final calculated radius of the circle is $\frac{5\sqrt{2}}{3}$. The final answer is $\boxed{\frac{5\sqrt{2}}{3}}$ which corresponds to option (B).