Finding the vector equation of a plane

[i_love_science]

Homework Statement:

Find a vector equation of the plane that passes through the point (6, 0, 0) and contains the line 𝑥 = 4 - 2𝑡, 𝑦 = 2 + 3𝑡, 𝑧 = 3 + 𝑡.

Relevant Equations:

vector equation

Solution:
u = [-2,3,1]
P_o = (6,0,0) & P = (4,2,3)
P_oP = v = [-2,2,3]
Therefore, the answer is [6,0,0] + r[-2,3,1] + q[-2,2,3]; r, q are real numbers

I don't understand why (6,0,0) is used as the point in the vector equation, since it only lies on the [-2,2,3] vector, not the u = [-2,3,1] vector.

My answer is r = [4,2,3] + r[2,-2,-3] + q[-2,3,1], r, q are real numbers

Could anyone explain the solution, and is my answer correct? Thanks.

 

[anuttarasammyak]

Your answer is transformable to
$$[4,2,3]+q[-2,3,1]-r[-2,2,3]=[6,0,0]+q[-2,3,1]-(r-1)[-2,2,3]$$
changing q with r and -(r-1) with q, it equals to the text answer.