### Three Dimensional Geometry Question 13

#### Question 13 - 29 January - Shift 1

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 $\alpha \in \mathbb{Z}$, if the area of $\triangle ABC$ is 21 sq. units and the line segment $BC$ has length $2 \sqrt{21}$ units, then $\alpha^{2}$ is equal to___________

## Show Answer

#### Answer: 9

#### Solution:

#### Formula: If two lines are perpendicular then dot product of their direction ratios will be zero (5.5), Area of triangle

A $(0, 2, \alpha)$

$|\frac{1}{2} \cdot 2 \sqrt{21} \cdot \begin{vmatrix} i & j & k \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{vmatrix} \frac{1}{\sqrt{25+4+9}}|=21 \sqrt{21}$

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

$\Rightarrow 12 \alpha^{2}+80 \alpha+450=798$

$\Rightarrow 12 \alpha^{2}+80 \alpha-348=0$

$\Rightarrow \alpha=3 \Rightarrow \alpha^{2}=9$