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

The plane contains the points $(-4,0,0),(0,3,0)$ and $(0,0,5)$

The plane that contains the points $(-4,0,0),(0,3,0)$ and $(0,0,5)$

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:

The equation of the plane satisfying the points $(-4,0,0),(0,3,0)$ and $(0,0,5)$ is obtained by substituting the points in $ax+by+cz=d$ and solve for the constants $a,b,c$ and $d$

The point $(-4,0,0)$ gives:

$a(-4)+b\left(0\right)+c\left(0\right)=d\left(Substitution\right)$

$-4a=d\left(Simplify\right)$

The point $(0,3,0)$ gives:

$a\left(0\right)+b\left(3\right)+c\left(0\right)=d$ (Substitution)

$b=\frac{d}{3}$ (Simplify)

The point $(0,0,5)$gives:

$$\begin{gathered} a\left(0\right)+b\left(0\right)+c\left(5\right)=d\text{(Substitution)} \\ c=\frac{d}{5}(\text{Simplify)} \end{gathered}$$

Substitute the values of constants $a,b$ and $c$ in the general form.

The equation of the plane is:

$-\frac{d}{4}x+\frac{d}{3}y+\frac{d}{5}z=d$ (Substitution)

$d\left(-\frac{x}{4}+\frac{y}{3}+\frac{z}{5}\right)=d$ (Take common factor out)

Thus, the equation of the plane that contains the points $(-4,0,0),(0,3,0)$ and $(0,0,5)$ is $-15x+20y+12z=60$