1. Key Concepts and Formulas

• Cramer's Rule for Infinitely Many Solutions: For a system of three linear equations $a_i x+b_i y+c_i z=d_i$ (for $i=1,2,3$) to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) and all the determinants formed by replacing a column of the coefficient matrix with the constant terms ($\Delta_x$, $\Delta_y$, $\Delta_z$) must all be zero. $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \quad \text{and} \quad \Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix} = 0, \quad \Delta_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix} = 0, \quad \Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} = 0$
• Distance from a Point to a Line: The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$
• Radius of a Tangent Circle: The radius of a circle tangent to a line is equal to the perpendicular distance from the center of the circle to that line.

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2. Step-by-Step Solution

Step 1: Set up the coefficient matrix and constant terms. We are given the system of linear equations: $\begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}$ From these equations, we can identify the coefficient matrix $A$ and the column vector of constant terms $D$: $A = \begin{pmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{pmatrix}, \quad D = \begin{pmatrix} 9 \\ 8 \\ 16-\mu \end{pmatrix}$

Step 2: Apply the condition $\Delta = 0$ to find $\lambda$. For the system to have infinitely many solutions, the determinant of the coefficient matrix, $\Delta$, must be zero. $\Delta = \begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{vmatrix}$ To calculate the determinant, we can expand along the first row: $\Delta = 2 \left( 3 \cdot (-(4+\lambda)) - (-2) \cdot 3 \right) - 3 \left( 7 \cdot (-(4+\lambda)) - (-2) \cdot 12 \right) + 5 \left( 7 \cdot 3 - 3 \cdot 12 \right)$ $\Delta = 2 \left( -12 - 3\lambda + 6 \right) - 3 \left( -28 - 7\lambda + 24 \right) + 5 \left( 21 - 36 \right)$ $\Delta = 2 \left( -6 - 3\lambda \right) - 3 \left( -4 - 7\lambda \right) + 5 \left( -15 \right)$ $\Delta = -12 - 6\lambda + 12 + 21\lambda - 75$ $\Delta = 15\lambda - 75$ Setting $\Delta = 0$ for infinitely many solutions: $15\lambda - 75 = 0$ $15\lambda = 75$ $\lambda = \frac{75}{15}$ $\lambda = 5$

Step 3: Apply the condition $\Delta_x = 0$ (or $\Delta_y$ or $\Delta_z$) to find $\mu$. Now that we have $\lambda = 5$, we substitute this value into the coefficient of $z$ in the third equation, which becomes $-(4+5) = -9$. The system of equations effectively becomes: $\begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-9 z=16-\mu \end{aligned}$ For infinitely many solutions, we must also have $\Delta_x = 0$. $\Delta_x$ is formed by replacing the first column of the coefficient matrix with the constant terms: $\Delta_x = \begin{vmatrix} 9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9 \end{vmatrix}$ Expanding along the first row: $\Delta_x = 9 \left( 3(-9) - (-2)(3) \right) - 3 \left( 8(-9) - (-2)(16-\mu) \right) + 5 \left( 8(3) - 3(16-\mu) \right)$ $\Delta_x = 9 \left( -27 + 6 \right) - 3 \left( -72 + (32 - 2\mu) \right) + 5 \left( 24 - (48 - 3\mu) \right)$ $\Delta_x = 9 \left( -21 \right) - 3 \left( -40 - 2\mu \right) + 5 \left( -24 + 3\mu \right)$ $\Delta_x = -189 + 120 + 6\mu - 120 + 15\mu$ $\Delta_x = -189 + 21\mu$ Setting $\Delta_x = 0$ for infinitely many solutions: $-189 + 21\mu = 0$ $21\mu = 189$ $\mu = \frac{189}{21}$ $\mu = 9$ Thus, the center of the circle is $(\lambda, \mu) = (5, 9)$.

Step 4: Calculate the radius of the circle. The circle is centered at $(\lambda, \mu) = (5, 9)$ and touches the line $4x = 3y$. First, rewrite the equation of the line in the standard form $Ax + By + C = 0$: $4x - 3y + 0 = 0$ Here, $A=4$, $B=-3$, and $C=0$. The center of the circle is $(x_0, y_0) = (5, 9)$. The radius $R$ of the circle is the perpendicular distance from its center to the tangent line. Using the distance formula: $R = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$ $R = \frac{|4(5) - 3(9) + 0|}{\sqrt{4^2 + (-3)^2}}$ $R = \frac{|20 - 27|}{\sqrt{16 + 9}}$ $R = \frac{|-7|}{\sqrt{25}}$ $R = \frac{7}{5}$

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3. Common Mistakes & Tips

• Determinant Calculation Errors: Be meticulous with signs and arithmetic when calculating determinants. A single sign error can lead to incorrect values for $\lambda$ and $\mu$. Expanding along a column with identical elements (like the second column here) can sometimes simplify calculation by factoring, but careful expansion is always key.
• Conditions for Infinitely Many Solutions: Remember that for an infinite number of solutions in a system of three linear equations, all four determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. If only $\Delta=0$ but one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.
• Line Equation Form: Ensure the line equation is in the $Ax+By+C=0$ form before applying the distance formula to avoid errors in identifying $A, B, C$. Forgetting the absolute value in the numerator of the distance formula is also a common mistake.

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4. Summary

This problem required a two-part approach. First, we utilized Cramer's Rule for systems of linear equations with infinitely many solutions to determine the unknown parameters $\lambda$ and $\mu$. By setting the determinant of the coefficient matrix ($\Delta$) to zero, we found $\lambda=5$. Then, by setting the determinant $\Delta_x$ (formed by replacing the x-coefficient column with constants) to zero, we found $\mu=9$. This established the center of the circle as $(5,9)$. In the second part, we used the geometric property that the radius of a circle tangent to a line is the perpendicular distance from its center to the line. Applying the distance formula from the center $(5,9)$ to the line $4x-3y=0$, we calculated the radius to be $\frac{7}{5}$.

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5. Final Answer

The radius of the circle is $\frac{7}{5}$. The final answer is $\boxed{\frac{7}{5}}$, which corresponds to option (A).