Parametric / vector question. Question 10 [10 points]
Let L be the line with parametric equations
$$ x = −6−3t $$
$$ y = 6+3t $$
$$ z = −8+2t $$
Find the vector equation for a line that passes through the point P=(−1, 2, 3) and intersects L at a point that is distance 2 from the point Q=(−6, 6, −8). Note that there are two possible correct answers.
If anyone already has a similar question up, even just pointing me in the right direction would help a lot!
 A: Hint: 
Note the point $Q$ is on $L$, so you just need to find the point $P'$ on $L$ s.t. $|P'Q|=2$, which means you just need to find corresponding $t$. There are two $t$ satisfying the requirement. 
After you get the points $P'$, just connect it with $P$ by a line.
A: Let the required line be L1
From the equation of line L: line L contains the point Q (t=0). A distance of 2 means it can happen in two ways. Imagine a circle of radius 2, with the origin set at point Q. Where this circle cuts the line L, will become the points of intersection. (This will happen exactly at 2 points, hence two answers)


*

*Find these points of intersection (co-ordinates) 
1a. Using distance formula, a distance of 2 from Q.
1b. This should form a quadratic equation in t. Solve it to get two values of t.
1c. Find the coordinates (x,y,z) corresponding to each value of t.

*Using these coordinates (one by one) and the coordinates of point P, find the equations of line L1.

A: This is my question:
http://s10.postimg.org/po9b9w5bt/Screenshot_2015_11_13_20_52_38.png
And Here is my answer:
http://s19.postimg.org/za2ihshfn/IMG_0356.jpg
Is it correct?
