Find the equations of the planes determined by the given conditions.

The plane contains the points $\left(x_0,0,0\right),\left(0,y_0,0\right)$ and $\left(0,0,z_0\right)$

The plane that contains the points $\left(x_0,0,0\right),\left(0,y_0,0\right)$ and $\left(0,0,z_0\right)$

The goal is to determine the plane equation that is determined by the given conditions.

The general form of the equation of the plane is:

$ax+by+cz=d$

The equation of the plane satisfying the points $\left(x_0,0,0\right),\left(0,y_0,0\right)$and $\left(0,0,z_0\right)$is obtained by substituting the points in $ax+by+cz=d$and solve for the constants $a,b,c$and $d$

The point $\left(x_0,0,0\right)$gives:

$$\begin{gathered} a\left(x_0\right)+b\left(0\right)+c\left(0\right)=d\text{(Substitution)} \\ a{x}_0=d(\text{Simplify)} \\ a=\frac{d}{x_0} \end{gathered}$$

The point $\left(0,y_0,0\right)$gives:

$a\left(0\right)+b\left(y_0\right)+c\left(0\right)=d$(Substitution)

$b=\frac{d}{y_0}\text{(Simplify)}$

The point $\left(0,0,z_0\right)$ gives:

In the general form, substitute the values of the constants $a,b$ and $c$

The equation of the plane is:

$\frac{d}{x_0}x+\frac{d}{y_0}y+\frac{d}{z_0}z=d$(Substitution)

$d\left(\frac{1}{x_0}x+\frac{1}{y_0}y+\frac{1}{z_0}z\right)=d$(Take common factor out)

$\frac{x}{x_0}+\frac{y}{y_0}+\frac{z}{z_0}=1$

$\frac{y_0{z}_{0}x+x_0{z}_{0}y+x_0{y}_{0}z}{x_0{y}_0{z}_0}=1$

$y_0{z}_{0}x+x_0{z}_{0}y+x_0{y}_{0}z=x_0{y}_0{z}_0$

Thus, the equation of the plane that contains the points $\left(x_0,0,0\right),\left(0,y_0,0\right)$and $\left(0,0,z_0\right)$is $y_0{z}_{0}x+x_0{z}_{0}y+x_0{y}_{0}z=x_0{y}_0{z}_0$