375. Find the distance:
1) from the point $(2 ; 3 ;-4)$ to the plane

$$
2 x+6 y-3 z+16=0
$$

2) from the point $(2 ;-4 ; 1)$ to the plane

$$
x-8 y+4 z=0
$$
Solution. 1) We will bring the equation to the normal form by multiplying the given equation by the normalizing factor

$$
M=\frac{1}{-\sqrt{4+36+9}}=-\frac{1}{7}
$$

We get

$$
-\frac{2}{7} x-\frac{6}{7} y+\frac{3}{7} z-\frac{16}{7}=0
$$

or

$$
\frac{2 x+6 y-3 z+16}{-7}=0
$$

The distance $d$ will be found using formula (15'):

$$
d=\left|\frac{2 \cdot 2+6 \cdot 3-3(-4)+16}{-7}\right|=\left|-\frac{50}{7}\right|=7 \frac{1}{7}
$$

Given that $D=16 \neq 0$, and $\delta=-\frac{50}{7}$, it follows that the given point and the origin are on the same side of the plane.
2) $M=\frac{1}{ \pm \sqrt{1+64+16}}= \pm \frac{1}{9}$ (the sign can be chosen arbitrarily, since the free term of the given equation $D=0$).

We will bring the equation to the normal form, i.e., multiply it by the normalizing factor $M$. We get

$$
\frac{x-8 y+4 z}{9}=0
$$

The distance

$$
d=\left|\frac{2-8(-4)+4 \cdot 1}{9}\right|=4 \frac{2}{9}
$$