Question

Find the vector equation of the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6). Also find the distance of point P(6, 5, 9) from this place.

Answer 1

$\left|\begin{array}{ccc}x-3& y+1& z-2\\ 5-3& 2+1& 4-2\\ -1-3& -1+1& 6-2\end{array}\right|=0$

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

$\Rightarrow (x-3)(12-0)-(y+1)(8+8)+(z-2)(0+12)=0$

$\Rightarrow 12x-36-16y-16+12z-24=0$

$\Rightarrow 12x-16y+12z-76=0$

$\Rightarrow 3x-4y+3z-19=0$

Now,
$\overrightarrow{r}.(3\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k})=19$

Now, distance of plane from point P(6, 5, 9)

$=\left|\frac{(3\times 6)+(-4)\left(5\right)+\left(3\right)\left(9\right)-19}{\sqrt{34}}\right|$

$=\left|\frac{18-20+27-19}{\sqrt{34}}\right|=\frac{6}{\sqrt{34}}$