<s>[INST] 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?
 [/INST] 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 $x=\frac{25}{3}$
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