Question

Find the equation of the plane passing through the points $(2,-3,1)$ and $(-1,1,-7)$ and perpendicular to the plane $x-2y+5z+1=0$

Answer 1

Required plane is perpendicular to the given plane

$x-2y+5z+1=0$

$\Rightarrow$ Required plane is parallel to the line which is perpendicular to the given plane.

Direction ratio of line $a=1,b=-2,c=5$.

Hence required plane is

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ a& b& c\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-2& y+3& z-1\\ -1-2& 1+3& -7-1\\ 1& -2& 5\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-2& y+3& z-1\\ -3& 4& -8\\ 1& -2& 5\end{array}\right|=0$

$\Rightarrow (x-2)(20-16)-(y+3)(-15+8)+(z-1)(6-4)=0$

$\Rightarrow (x-2)4-(y+3)(-7)+(z-1)2=0$

$\Rightarrow 4x+7y+2z+11=0$