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QJEEM23Ma7462: JEE Main 2023 Mathematics - Class 12 JEEM23Ma7462 Solution: Let the co-ordinates of one vertex of be and the other two…

Let the co-ordinates of one vertex of ΔABC\Delta ABC be A(0,2,α)A(0,2,\alpha) and the other two vertices lie on the line x+α5=y−12=z+43{{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3} . For α∈Z\alpha \in \mathbb{Z} , if the area of ΔABC\Delta ABC is 21 sq. units and the line segment BCBC has length 2212\sqrt{21} units, then α2\alpha^2 is equal to ___________.

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A. (O12,α)\left(\mathrm{O}_{1} 2, \alpha\right)

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - 3D Geometry Question 110 English Explanation

∣12⋅221.∣iamp;jamp;kαamp;1amp;α+45amp;2amp;3∣125+4+9∣=2121\left|\frac{1}{2} \cdot 2 \sqrt{21}.\right| \begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{array}\left|\frac{1}{\sqrt{25+4+9}}\right|=21 \sqrt{21}

(2α+5)2+(2α+20)2+(2α−5)2=2138\sqrt{(2 \alpha+5)^{2}+(2 \alpha+20)^{2}+(2 \alpha-5)^{2}}=\sqrt{21} \sqrt{38}

⇒12α2+80α+450=798\Rightarrow 12 \alpha^{2}+80 \alpha+450=798

⇒12α2+80α−348=0\Rightarrow 12 \alpha^{2}+80 \alpha-348=0

⇒α=3⇒α2=9\Rightarrow \alpha=3 \Rightarrow \alpha^{2}=9

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