Determine points on line with specific distance from plane There is a line $$p: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1}$$
and plane $$\pi : x+y+2z-3=0.$$
I need to find points $T_1, T_2 \in p$. Requirement when finding those points are that they have a set distance from plane $\pi$, with distance being $\sqrt{6}$, and that the points are on the line.
The solution for this problem is
$T_1=\left( \dfrac 13, -2, -\dfrac 23\right)$
$T_2 = \left( \dfrac{25}{3}, 10, -\dfrac{14}{3} \right)$
Problem is I don't have idea how to get there, so I will apreciate any advice/guidance for solving this problem.
EDIT1
Equation for distance between point and plane is:  
$d(pi,T_\pi')=\dfrac{|AX_0+BY_0+CZ_0+D|}{\sqrt{(A^2+B^2+C^2)}})$
so I get:
$\sqrt6=\dfrac{|X_0+Y_0+2Z_0-3|}{\sqrt6}$
$6=|X_0+Y_0+2Z_0-3|$
since the requirement for parallel planes is: $\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}$ I deduced that $A_2=B_2=2C_2$
returning to where I left:
$6=|4X_0-3|$
$4X_0-3=6$
and
$4X_0-3=-6$
$X_01=\dfrac{9}{4}$
and
$X_02=\dfrac{-3}{4}$
Finally I have two planes but I will focus on one because I messed up something:
$\pi .. \dfrac{9}{4}X+\dfrac{9}{4}Y+\dfrac{9}{2}-3=0$
If I fit vector of line direction into this equation we get that $t=\dfrac{10}{9}$
Now if try to fit $t$ back in the equation I will get wrong results since x will be $\dfrac{29}{9}$. 
What am I currently missing?

If you are going to work with parallel planes, you can use the fact that they have the same normal vector. This means you only need to change the $d$ in the plane equation. So you want to find a plane $x+y+2z+d=0$ such that its distance to $\pi$ is $\sqrt{6}$.
The distance between two planes with the same normal vector is:
$ d=\frac{|d_1-d_2|}{\sqrt{A^2+B^2+C^2}}$
So, plugging in the values for this specific problem you get:
$ \sqrt{6}=\frac{|d_1-(-3)|}{\sqrt{6}}=\frac{|d_1+3|}{\sqrt{6}}$
So either $d=-9$ or $d=3$
So, for example, your first plane would be:
$x+y+2z-9=0$
Then,  fitting the vector of line direction into this equation we get that $t=\frac{11}{3}$
Then you get $x=\frac{25}{3}$
