Question

Find the distance from the plane passing through $\left(1,3,2\right),\left(-5,0,2\right),\left(1,1,-4\right)$ to the point $(2,3,4)$

Answer 1

Let $(x_1,y_1,z_1)\equiv (1,3,2)$

$(x_2,y_2,z_2)\equiv (-5,0,2)$

$(x_3,y_3,z_3)\equiv (1,1,-4)$

So, equation of plane will be

$|x-x_{1}y-y_{1}z-z_1{x}_2-x_1{y}_2-y_1{z}_2-z_1{x}_3-x_1{y}_3-y_1{z}_3-z_1|=0$

$|x-1y-3z-2-6-300-2-6|=0$

$(x-1)\left(18\right)-(y-3)\left(36\right)+(z-2)\left(12\right)=0$

$3(x-1)-6(y-3)+2(z-2)=0$

$3x-6y+2z+11=0$

distance from $(2,3,4)=d=\left|\frac{3\left(2\right)-6\left(3\right)+2\left(4\right)+11}{\sqrt{3^2+{(-6)}^2+{\left(2\right)}^2}}\right|$

$d=\frac{7}{7}=1$