Line and point in plane

$\frac{x-2}{3}=-\frac{y}{4}=\frac{1}{5} (-z-2)$ with $y\to 0$ gives $x\to 2,z\to -2$

$\frac{x-2}{3}=-\frac{y}{4}=\frac{1}{5} (-z-2)$ with $y\to 2$ gives $x\to \frac12,z\to \frac12$

$\text{p0}=\{5,-2,7\}$ , $\text{p1}=\{2,0,-2\}$ and $\text{p2}=\left\{\frac{1}{2},2,\frac{1}{2}\right\}$ are three points on the plane.

The unit normal to the plane is $\text{Normalize}[(\text{p1}-\text{p0})\times (\text{p2}-\text{p0})] \Rightarrow \left\{\frac{23}{\sqrt{979}},\frac{21}{\sqrt{979}},-\frac{3}{\sqrt{979}}\right\}$ .

The minimum distance from the origin, $\left\{\frac{23}{\sqrt{979}},\frac{21}{\sqrt{979}},-\frac{3}{\sqrt{979}}\right\}.\text{p0}$ , is $\frac{52}{\sqrt{979}}$ . Actually, any point on the plane could have been used. $\text{p0}$ was chosen.

Therefore the plane's equation modulo any multiplicative factor is $\left\{\frac{23}{\sqrt{979}},\frac{21}{\sqrt{979}},-\frac{3}{\sqrt{979}}\right\}.\{x,y,z\}-\frac{52}{\sqrt{979}}=0$ .

Multiplying away the common denominator and writing the plane's equation in the specified form: $23 x+21 y-3 z=52$

All three points checked as being on the plane. That work not shown.

$\left\lfloor -\frac{23\ 21}{3\ 52}\right\rfloor \Rightarrow -4$ .