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  • Let the co-ordinates of one vertex of <img alt="\mathrm{\triangle A B C \: be \: A(0,2, \alpha)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctriangle%20A%20B%20C%20%5C%3A%2

Let the co-ordinates of one vertex of \mathrm{\triangle A B C \: be \: A(0,2, \alpha)} and the other two vertices lie on the line \mathrm{\frac{x+\alpha}{5}=\frac{y-1}{2}=\frac{z+4}{3}}. For \mathrm{\alpha \in \mathbb{Z},} if the area of  \mathrm{\triangle A B C \: is \: 21 \mathrm{sq}.,} units and the line segment \mathrm{B C} has length \mathrm{2 \sqrt{21}} units, then \mathrm{\alpha^2} is equal to

Option: 1

9


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{A\left ( 0,2,\alpha \right )}

\begin{aligned} & \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{(2 \alpha+5)^2+(2 \alpha+20)^2+(2 \alpha-5)^2}=\sqrt{21} \sqrt{38} \\ & 12 \alpha^2+80 \alpha+450=798 \\ & 12 \alpha^2+80 \alpha-398=0 \\ & \alpha=3 \Rightarrow \alpha^2=9 \end{aligned}

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Gautam harsolia

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