Question:

<p>Let the tangent lengths from vertices <i>A</i>, <i>B</i>, <i>C</i> to the incircle be <i>s-a</i> = 4, <i>s-b</i> = 6, <i>s-c</i> = 8.</p><p>Then: \(s - a = 4\), \(s - b = 6\), \(s - c = 8\)</p><p>Adding: \(3s - (a+b+c) = 18\)</p><p>Since \(s = \frac{a+b+c}{2}\), we have \(3s - 2s = 18\), so \(s = 18\)</p><p>Therefore: \(a = 14\), \(b = 12\), \(c = 10\)</p><p>Check: \(a + c = 14 + 10 = 24 = 2 \times 12 = 2b\), so \(a, b, c\) are in AP. ✓</p><p>Area by Heron's formula: \(K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18 \times 4 \times 6 \times 8} = \sqrt{3456} = 12\sqrt{24} = 24\sqrt{6}\)</p><p>For exradii: \(r_1 = \frac{K}{s-a} = \frac{24\sqrt{6}}{4} = 6\sqrt{6}\), \(r_2 = \frac{24\sqrt{6}}{6} = 4\sqrt{6}\), \(r_3 = \frac{24\sqrt{6}}{8} = 3\sqrt{6}\)</p><p>Check if in HP: \(\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}\) should be in AP: \(\frac{1}{6\sqrt{6}}, \frac{1}{4\sqrt{6}}, \frac{1}{3\sqrt{6}}\). Yes, they form an AP. ✓</p><p>Options (b) and (c) are correct.</p>