<p><strong>74 (B):</strong> Let a, b, c ∈ ℝ<sup>+</sup> and the system of equations<br/>\[(1-a)x + y + z = 0\]<br/>\[x + (1-b)y + z = 0\]<br/>\[x + y + (1-c)z = 0\]<br/>has infinitely many solutions. If λ be the minimum value of abc, then λ is divisible by</p>
Step-by-Step Solution
Key Concept: The system has infinitely many solutions when det = 0. Use this condition to derive a constraint on a, b, c, then apply AM-GM to find the minimum of abc.
<p><strong>Solution:</strong></p><p>For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero:</p><p>$$\begin{vmatrix} 1-a & 1 & 1 \\ 1 & 1-b & 1 \\ 1 & 1 & 1-c \end{vmatrix} = 0$$</p><p>Expanding and simplifying, this condition gives:</p><p>$$abc = ab + bc + ca$$</p><p>By AM-GM inequality: $$\frac{ab + bc + ca}{3} \geq \sqrt[3]{(abc)^2}$$</p><p>Since abc = ab + bc + ca, we have $$\frac{abc}{3} \geq \sqrt[3]{(abc)^2}$$</p><p>Let abc = λ. Then λ ≥ 27, with minimum λ = 27.</p><p>Therefore, λ is divisible by 3, 9, and 27. Answer: 6 (or 3, 9 depending on options).</p>
Correct Answer: 6