Question

Find the equation of the plane determined by the points $A(3,-1,2),B(5,2,4)$ and $C(-1,-1,6)$ and hence find the distance between the plane and the point $P(6,5,9)$

Answer 1

The equation of the plane through three non-collinear points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6) can be expressed as
$\left|\begin{array}{ccc}x-3& y+1& z-2\\ 5-3& 2+1& 4-2\\ -1-3& -1+1& 6-2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x-3& y+1& z-2\\ 2& 3& 2\\ -4& 0& 4\end{array}\right|=0$
$\Rightarrow 12(x-3)-16(y+1)+12(z-2)=0$
$\Rightarrow 12x-16y+12z-76=0$
$\Rightarrow 3x-4y+3z-19=0$ is the required equation.
Now, distance of P(6, 5, 9) from the plane is given by
$=\left|\frac{3\times 6-4\left(5\right)+3\left(9\right)-19}{\sqrt{9+16+9}}\right|=\left|\frac{6}{\sqrt{34}}\right|$ sq.units.