Question

The equation of plane passing through $A(2,2,-1),B(3,4,2)$ and $C(7,0,6)$ is also passing through $D(1,\alpha ,2\beta )$ such that $OD=\sqrt{46}.$ Then the point $D$ with integral coordinates is:

• A. $(1,6,-3)$
• B. $(1,3,6)$
• C. $(1,-3,-6)$
• D. $(1,3,-6)$

Answer 1

The correct option is C $(1,-3,-6)$

Let the three given points be $(x_1,y_1,z_1)=(2,2,-1),(x_2,y_2,z_2)=(3,4,2)$ and $(x_3,y_3,z_3)=(7,0,6)$
Then the equation of the plane is:
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x-2& y-2& z-(-1)\\ 3-2& 4-2& 2-(-1)\\ 7-2& 0-2& 6-(-1)\end{array}\right|=0$
On simplifying, we have $5x+2y-3z=17$
Now this plane passes through $(1,\alpha ,2\beta )\Rightarrow \alpha -3\beta =6$
So the point $D=(1,6+3\beta ,2\beta )$
Given $OD=\sqrt{46}\Rightarrow 1+(6+3\beta {)}^2+(2\beta {)}^2=46\Rightarrow \beta =-3\text{or}\beta =\frac{3}{13}$
$\therefore D=(1,-3,-6)$(with integral co-ordinates)