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?

Let $ \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1} = t$ for some point $(x,y,z)$ on the line $p$.
Then $p(t) = (1+2t, -1+3t, -1-t)$
Using your distance formula, we get
\begin{align}
   \dfrac{|1(1+2t) + 1(-1+3t) + 2(-1-t) - 3|}{\sqrt{(1^2 + 1^2 + 2^2)}} &= \sqrt 6 \\
   |3t-5| &=  6 \\
   3t - 5 &= \pm 6 \\
   t &\in \left\{ -\dfrac 13, \dfrac{11}{3} \right\}
\end{align}
So the points are
$p\left(-\dfrac 13 \right) = T_1$
$p\left( \dfrac{11}{3} \right) = T_2$
