4.4. Given the equation of a plane in general form $2 x-2 y+$ $+z-20=0$.

Transform this equation: 1) into the equation of a plane in intercept form; 2) into the normal equation of the plane.
Solution.

1) Let's calculate the lengths of the segments cut off by the plane on the coordinate axes, counting from the origin. Assuming $y_{0}=0, z_{0}=0$, we get $2 x_{0}-2 \cdot 0+0-20=0, x_{0}=10=a$. Now let $x_{1}=0, z_{1}=0$, then $2 \cdot 0-2 \cdot y_{1}+0-20=0$, $y_{1}=-10=b$. Finally, we find the segment cut off by the plane on the $O Z$ axis, for which we set $x_{2}=0, y_{2}=0,2 \cdot 0-$ $-2 \cdot 0+z_{2}-20=0, z_{2}=20=c$. The equation of the plane in intercept form is

$$
x / 10+y /(-10)+z / 20=1
$$

2) To write the normal equation of the plane, we calculate

$$
\sqrt{A^{2}+B^{2}+C^{2}}=\sqrt{2^{2}+(-2)^{2}+1^{2}}=3
$$

Since the sign of the free term is negative, the normal equation is

$$
-2 / 3 \cdot x+2 / 3 \cdot y-1 / 3 \cdot z+20 / 3=0
$$