Find the unique line–plane crossing

Let the intersection be B. Since B lies on the line, its position is $(4+\lambda,-6-2\lambda,-2)$. Since it also lies on the plane,

$\begin{aligned}2(4+\lambda)-(-6-2\lambda)+2(-2)&=4,\\8+2\lambda+6+2\lambda-4&=4,\\10+4\lambda&=4,\\\lambda&=-\frac32.\end{aligned}$

Substitute into the line:

$\begin{aligned}\overrightarrow{OB}&=(4-\tfrac32,-6+3,-2)\\&=(\tfrac52,-3,-2).\end{aligned}$

The direction–normal product is $(1,-2,0)\cdot(2,-1,2)=4\ne0$, so this is the unique crossing. As an independent check, the point gives $2(5/2)-(-3)+2(-2)=4$. The source names the intersection B; that label is retained.