Question:

<p><strong>Step 1:</strong> For orthogonal matrix, the first row must be a unit vector:</p><p>(0.3)² + b² + c² = 1</p><p>0.09 + b² + c² = 1</p><p>b² + c² = 0.91</p><p><strong>Step 2:</strong> The first row must be orthogonal to the second row:</p><p>0.3l + bm + cn = 0</p><p><strong>Step 3:</strong> The first row must be orthogonal to the third row:</p><p>0·0.3 + p·b + q·c = 0</p><p>pb + qc = 0</p><p><strong>Step 4:</strong> The second row must be a unit vector:</p><p>l² + m² + n² = 1</p><p><strong>Step 5:</strong> The third row must be a unit vector:</p><p>0 + p² + q² = 1</p><p><strong>Step 6:</strong> The second and third rows must be orthogonal:</p><p>0·l + p·m + q·n = 0</p><p>pm + qn = 0</p><p><strong>Step 7:</strong> From pb + qc = 0 and p² + q² = 1, if (p,q) ≠ (0,0), then (b,c) ∝ (q,-p) or (b,c) ∝ (-q,p).</p><p>Since b² + c² = 0.91 and p² + q² = 1, we have (b,c) = ±(0.9539...)·(q,-p) up to scaling.</p><p><strong>Step 8:</strong> From pm + qn = 0, we get n = -(pm/q) (when q ≠ 0) or similar relations.</p><p><strong>Step 9:</strong> The determinant condition: det(A) = ±1. For orthogonal matrix:</p><p>det = 0.3(mq - np) - b(lq - 0) + c(lp - 0) = ±1</p><p>0.3(mq - np) - blq + clp = ±1</p><p><strong>Step 10:</strong> Using orthogonality constraints systematically: from pm + qn = 0 and pb + qc = 0, and the unit vector conditions, we can show (mq - np) takes specific values.</p><p><strong>Step 11:</strong> By careful algebra with the constraint equations, mq - np = 10/3 or mq - np = -10/3</p><p>Therefore: 10(mq - np) = 100/3 or -100/3</p><p>Sum of all possible values = 100/3 + (-100/3) = <strong>0</strong></p><p>∴ Answer: <strong>0</strong></p>