Finding the vector equation of a plane

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The discussion centers on finding the vector equation of a plane using specific points and direction vectors. The initial solution presents the equation as [6,0,0] + r[-2,3,1] + q[-2,2,3], where (6,0,0) is questioned as a point on the plane. A participant suggests an alternative point (4,2,3) and proposes a different equation format. The conversation highlights the need for clarification on the choice of point and the correctness of the alternative solution. The thread emphasizes understanding vector equations in relation to given points and direction vectors.
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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]
Po = (6,0,0) & P = (4,2,3)
PoP = 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.
 
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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.
 
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