Question:

<p><strong>Step 1: Set up the system in matrix form</strong></p><p>The given equations are:</p><p>aλ + bm + cn = 0</p><p>bλ + cm + an = 0</p><p>cλ + am + bn = 0</p><p>This can be written as:</p><p>$$\begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \begin{pmatrix} λ \\ m \\ n \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$</p><p><strong>Step 2: Apply condition for non-trivial solution</strong></p><p>Since λ, m, n ∈ ℝ - {0}, we need a non-trivial solution. This requires:</p><p>$$\det\begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} = 0$$</p><p><strong>Step 3: Calculate the determinant</strong></p><p>Expanding the determinant:</p><p>$$\det = a(cb - a²) - b(b² - ac) + c(ab - c²)$$</p><p>$$= abc - a³ - b³ + abc + abc - c³$$</p><p>$$= 3abc - a³ - b³ - c³$$</p><p><strong>Step 4: Use the factorization</strong></p><p>We know that: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)</p><p>Therefore: 3abc - a³ - b³ - c³ = -(a + b + c)(a² + b² + c² - ab - bc - ca)</p><p>For this to equal zero (with a, b, c distinct):</p><p>$$a + b + c = 0$$</p><p><strong>Step 5: Calculate f(1)</strong></p><p>Given f(x) = ax³ + bx² + cx + 2</p><p>$$f(1) = a(1)³ + b(1)² + c(1) + 2$$</p><p>$$f(1) = a + b + c + 2$$</p><p>Since a + b + c = 0:</p><p>$$f(1) = 0 + 2 = 2$$</p><p><strong>∴ Answer: 2</strong></p>