Question:

<p><strong>Solution:</strong> Given \(a < 0, b < 0, c < 0\) and \(a + b + c = abc\)</p><p>Since all are negative, let \(a = -p, b = -q, c = -r\) where \(p, q, r > 0\).</p><p>Then \(-p - q - r = -pqr\), so \(p + q + r = pqr\)</p><p>By AM-GM inequality: \(\dfrac{p+q+r}{3} \geq \sqrt[3]{pqr}\)</p><p>This gives \(\dfrac{pqr}{3} \geq \sqrt[3]{pqr}\), which implies \(pqr \geq 3\sqrt{3}\)</p><p>At least one of \(p, q, r\) must exceed \(\dfrac{3}{2}\), hence at least one of \(a, b, c\) exceeds \(-\dfrac{3}{2}\) or is less than \(-\dfrac{3}{2}\).</p><p>∴ Answer is (a).</p>