1. Key Concepts and Formulas

• System of Linear Equations: For a system of three linear equations in three variables ($x, y, z$): $a_1 x + b_1 y + c_1 z = d_1$ $a_2 x + b_2 y + c_2 z = d_2$ $a_3 x + b_3 y + c_3 z = d_3$
• Conditions for Infinitely Many Solutions (Cramer's Rule): A system of linear equations has infinitely many solutions if and only if the determinant of the coefficient matrix ($\Delta$) is zero, AND all the auxiliary determinants ($\Delta_x, \Delta_y, \Delta_z$) are also zero.
  • $\Delta = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| = 0$
  • $\Delta_x = \left|\begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array}\right| = 0$
  • $\Delta_y = \left|\begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array}\right| = 0$
  • $\Delta_z = \left|\begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array}\right| = 0$
• Determinant of a $3 \times 3$ Matrix: The determinant of $\left|\begin{smallmatrix} a & b & c \\ d & e & f \\ g & h & i \end{smallmatrix}\right|$ is $a(ei-fh) - b(di-fg) + c(dh-eg)$.
• Algebraic Identity: $(A+B)^2 + (A-B)^2 = 2(A^2+B^2)$.

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2. Step-by-Step Solution

We are given the system of equations:

1. $2 x+y-z=5$
2. $2 x-5 y+\lambda z=\mu$
3. $x+2 y-5 z=7$

Step 1: Calculate the coefficient determinant ($\Delta$) and find $\lambda$.

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero ($\Delta = 0$). First, let's form the coefficient matrix and calculate its determinant. The coefficients are: $a_1=2, b_1=1, c_1=-1$ $a_2=2, b_2=-5, c_2=\lambda$ $a_3=1, b_3=2, c_3=-5$

$\Delta = \left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|$ Expanding along the first row: $\Delta = 2 \cdot \left| \begin{array}{cc} -5 & \lambda \\ 2 & -5 \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 2 & \lambda \\ 1 & -5 \end{array} \right| + (-1) \cdot \left| \begin{array}{cc} 2 & -5 \\ 1 & 2 \end{array} \right|$ $\Delta = 2((-5)(-5) - (\lambda)(2)) - 1((2)(-5) - (\lambda)(1)) - 1((2)(2) - (-5)(1))$ $\Delta = 2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5)$ $\Delta = 50 - 4\lambda + 10 + \lambda - 9$ $\Delta = (50 + 10 - 9) + (-4\lambda + \lambda)$ $\Delta = 51 - 3\lambda$

Now, apply the condition $\Delta = 0$: $51 - 3\lambda = 0$ $3\lambda = 51$ $\lambda = \frac{51}{3}$ $\lambda = 17$

Step 2: Calculate an auxiliary determinant ($\Delta_z$) and find $\mu$.

Since the system has infinitely many solutions, not only must $\Delta=0$, but also $\Delta_x=0$, $\Delta_y=0$, and $\Delta_z=0$. We can use any one of these to find $\mu$. Let's use $\Delta_z = 0$. The constant terms are: $d_1=5, d_2=\mu, d_3=7$. The auxiliary determinant $\Delta_z$ is formed by replacing the third column (coefficients of $z$) in $\Delta$ with the constant terms: $\Delta_z = \left|\begin{array}{ccc} 2 & 1 & 5 \\ 2 & -5 & \mu \\ 1 & 2 & 7 \end{array}\right|$ Expanding $\Delta_z$ along the first row: $\Delta_z = 2 \cdot \left| \begin{array}{cc} -5 & \mu \\ 2 & 7 \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 2 & \mu \\ 1 & 7 \end{array} \right| + 5 \cdot \left| \begin{array}{cc} 2 & -5 \\ 1 & 2 \end{array} \right|$ $\Delta_z = 2((-5)(7) - (\mu)(2)) - 1((2)(7) - (\mu)(1)) + 5((2)(2) - (-5)(1))$ $\Delta_z = 2(-35 - 2\mu) - 1(14 - \mu) + 5(4 + 5)$ $\Delta_z = -70 - 4\mu - 14 + \mu + 5(9)$ $\Delta_z = -70 - 4\mu - 14 + \mu + 45$ $\Delta_z = (-70 - 14 + 45) + (-4\mu + \mu)$ $\Delta_z = (-84 + 45) - 3\mu$ $\Delta_z = -39 - 3\mu$

Now, apply the condition $\Delta_z = 0$: $-39 - 3\mu = 0$ $-3\mu = 39$ $\mu = \frac{39}{-3}$ $\mu = -13$

Step 3: Calculate the required expression $(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$.

We have found $\lambda = 17$ and $\mu = -13$. The expression to evaluate is $(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$.

Substitute the values: $(17 + (-13))^{2} + (17 - (-13))^{2}$ $= (17 - 13)^{2} + (17 + 13)^{2}$ $= (4)^{2} + (30)^{2}$ $= 16 + 900$ $= 916$

Alternatively, using the algebraic identity $(A+B)^2 + (A-B)^2 = 2(A^2+B^2)$: $(\lambda+\mu)^{2}+(\lambda-\mu)^{2} = 2(\lambda^2 + \mu^2)$ $= 2(17^2 + (-13)^2)$ $= 2(289 + 169)$ $= 2(458)$ $= 916$

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3. Common Mistakes & Tips

• Sign Errors in Determinant Expansion: Be extremely careful with the alternating signs and arithmetic when calculating $3 \times 3$ and $2 \times 2$ determinants. This is a frequent source of error.
• Misapplication of Cramer's Rule Conditions: Remember that for infinitely many solutions, all determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. If $\Delta=0$ but any $\Delta_i \neq 0$, there is no solution.
• Leverage Algebraic Identities: Using identities like $(a+b)^2+(a-b)^2 = 2(a^2+b^2)$ can simplify calculations and save time in the final step.

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4. Summary

This problem required a thorough understanding and application of Cramer's Rule for systems of linear equations. By correctly identifying the conditions for infinitely many solutions ($\Delta=0$ and all auxiliary determinants also zero), we systematically calculated the determinant of the coefficient matrix to find $\lambda$, and then used one of the auxiliary determinants to find $\mu$. Finally, we substituted these values into the given expression, using an algebraic identity for efficiency, to arrive at the solution.

The final answer is $\boxed{\text{916}}$, which corresponds to option (A).