Question

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

Answer 1

$ \left|\frac{1}{2} \cdot 2 \sqrt{21} \begin{vmatrix} i & j & k \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{vmatrix} \frac{1}{\sqrt{25+4+9}}\right|=21 \sqrt{21} $
$ \sqrt{(2 \alpha+5)^2+(2 \alpha+20)^2+(2 \alpha-5)^2}=\sqrt{21} \sqrt{38} $
$\Rightarrow 12 \alpha^2+80 \alpha+450=798 $
$ \Rightarrow 12 \alpha^2+80 \alpha-348=0$
$ \Rightarrow \alpha=3 \Rightarrow \alpha^2=9$