Question:
Let \(\alpha_1, \alpha_2\) be the values of \(\alpha\) such that the distance between the point \((2, 4, 3)\) and the plane \(3x + y + \alpha z + 10 = 0\) is \(\sqrt{35}\) units. Then the area of the triangle with vertices \((\alpha_1, \alpha_2, 0)\), \((\alpha_2, \alpha_1, 0)\) and \(\left(\frac{164}{13}, 5, 0\right)\) is ________ unit\(^2\).
Let \(\alpha_1, \alpha_2\) be the values of \(\alpha\) such that the distance between the point \((2, 4, 3)\) and the plane \(3x + y + \alpha z + 10 = 0\) is \(\sqrt{35}\) units. Then the area of the triangle with vertices \((\alpha_1, \alpha_2, 0)\), \((\alpha_2, \alpha_1, 0)\) and \(\left(\frac{164}{13}, 5, 0\right)\) is ________ unit\(^2\).
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Use the point-to-plane distance formula to get a quadratic in \(\alpha\); then compute the area of the planar triangle in the plane \(z=0\).
Updated On: Nov 12, 2026
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Correct Answer: 35
Solution and Explanation
Distance: \(\dfrac{|3(2)+4+3\alpha+10|}{\sqrt{9+1+\alpha^2}}=\sqrt{35}\Rightarrow(3\alpha+20)^2=35(\alpha^2+10)\).
\(9\alpha^2+120\alpha+400=35\alpha^2+350\Rightarrow 26\alpha^2-120\alpha-50=0\Rightarrow13\alpha^2-60\alpha-25=0\).
\(\alpha=\dfrac{60\pm\sqrt{3600+1300}}{26}=\dfrac{60\pm70}{26}\), so \(\alpha=5\) or \(-\dfrac{5}{13}\).
Take \(A(5,-\tfrac{5}{13})\), \(B(-\tfrac{5}{13},5)\), \(C(\tfrac{164}{13},5)\). Points B and C have the same \(y=5\), so \(BC=\dfrac{164}{13}+\dfrac{5}{13}=13\). The height from A is \(5+\dfrac{5}{13}=\dfrac{70}{13}\).
Area \(=\dfrac12\times13\times\dfrac{70}{13}=35\).
Answer: 35
\(9\alpha^2+120\alpha+400=35\alpha^2+350\Rightarrow 26\alpha^2-120\alpha-50=0\Rightarrow13\alpha^2-60\alpha-25=0\).
\(\alpha=\dfrac{60\pm\sqrt{3600+1300}}{26}=\dfrac{60\pm70}{26}\), so \(\alpha=5\) or \(-\dfrac{5}{13}\).
Take \(A(5,-\tfrac{5}{13})\), \(B(-\tfrac{5}{13},5)\), \(C(\tfrac{164}{13},5)\). Points B and C have the same \(y=5\), so \(BC=\dfrac{164}{13}+\dfrac{5}{13}=13\). The height from A is \(5+\dfrac{5}{13}=\dfrac{70}{13}\).
Area \(=\dfrac12\times13\times\dfrac{70}{13}=35\).
Answer: 35
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