#### Answer

$(\dfrac{4}{3},\dfrac{-2}{3},\dfrac{-2}{3})$

#### Work Step by Step

As we given are that the parametric equations are: $x=2t; y=-t; z=-t$
The equation of a normal vector to the plane is: $n=\lt 2,-1,-1 \gt$
this implies, $3(2t)-5(-t)+2(-t)=6$
or, $9t =6 \implies t=\dfrac{2}{3}$
and $x=2(\dfrac{2}{3})=\dfrac{4}{3}; y=-\dfrac{2}{3}; z=-\dfrac{2}{3}$
Hence, the line will meet the plane at the point: $(\dfrac{4}{3},\dfrac{-2}{3},\dfrac{-2}{3})$