How can I find the line that contains the point $\ (0,2,1) $ and intersect the line $\ (2t,1-t,2+t) $ in $\ 90 $ degrees
Maybe since the direction vector of the given line is $\ (2,-1,1) $ and the the direction vector of the line I look for is perpendicular, the dot product of both should be zero?
The line that you're after must be contained in the plane $P$ defined by $2x-y+z=-1$, since this is the plane which is orthogonal to the given line (let us call it $l$) and which passes through $(0,2,1)$. Now, let us find $P\cap l$: we are after a number $t$ such that $2(2t)-(1-t)+2+t=-1$. But$$2(2t)-(1-t)+2+t=-1\iff6t=-2\iff t=-\frac13.$$So, the line that you're after is the line defined by the points $(0,2,1)$ and $\left(-\frac23,\frac43,\frac53\right)$.
An alternaive approach which does not use plane equations is this: consider the point $(0,1,2)\in l$. You are after a number $t\neq0$ such that the angle between $(0,1,2)$, $(2t,1-t,2+t)$, and $(0,2,1)$ is a right angle. This is equivalent to$$\bigl\langle(2t,1-t,2+t)-(0,1,2),(2t,1-t,2+t)-(0,2,1)\bigr\rangle=0.$$The only non-zero solution of this equation is $t=-\frac13$. So, again, the line that you are after is the line defined by the points $(0,2,1)$ and $\left(-\frac23,\frac43,\frac53\right)$.