Let a line with direction ratios a, 4a, 7 be perpendicular to the lines with direction ratios 3, 1, 2b and b, a, 2. If the point of intersection of the line $\frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac{z}{1}$ and the plane x – y + z = 0 is $\left(\alpha ,\beta ,\gamma \right),then\alpha +\beta +\gamma$ is equal to………

<p> (a, 4a, 7). (3, 1, 2b)= 0</p><p>3a + 4a – 14b = 0 a – 2b = 0 ……. (i)</p><p> (a, 4a, 7) . (b, a, 2) = 0</p><p>ab – 4a² + 14 = 0……… (ii)</p><p> (i) &amp; (ii) =&gt;2b² – 16b²+ 14 = 0</p><p><math><mrow><mo>⇒</mo><msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>1</mn><mo>⇒</mo><mi>a</mi><mo>=</mo><mn>2</mn><mi>b</mi><mo>=</mo><mo>±</mo><mn>2</mn></mrow></math></p><p>Plane : x – y + z = 0</p><p>P (4, 5, 1)</p><p>4 + 5 + 1 = 10</p>

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