Math: Equations of Lines

Determine the parametric equations of each of the following planes:

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g) the plane containing the 2 intersecting lines:

r = (5,4,2) + t(4,-2,1)
r = (1,6,-6) + s(6,-4,4)

My Attempt:
x = 5 + 4t + 6s
y = 4 -2t - 4s
z = 2 + t + 4s

####What did I do wrong in this question? Please point out and correct my mistake!

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h)the plane containing the point (8,3,5) and the line:

r = (1,3,-1) + t(2,2,-5)

My Attempt:
x = 8 + 2t
y = 3 + 2t
z = 5 - 5t

####How do I find the other direction vector for this question?

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  1. I can not find values of s and t to make those two lines intersect.
    At the intersection, the x, y and z values must be the same but
    for x : 5 + 4 t = 1 + 6 s
    for y : 4 - 2 t = 6 - 4 s
    for z : 2 + 1 t = -6 + 4 s

    In the second problem find the equation of a vector through the point perpendicular to the first line ( the line better not go through (8,3,5) )
    the plane will then contain those two lines

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  2. Oh or more easily in the second problem, a line containing (1,3,-1) where your given vector starts and (8,3,5) lies in the plane so you could use it as the second vector.

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  3. h)the plane containing the point (8,3,5) and the line:

    r = (1,3,-1) + t(2,2,-5)
    first find the vector from (1,3,-1) to (8,3,5)
    7 i + 0 j + 6 k
    the given vector has components
    2 i + 2 j -5 k
    the cross (vector ) product is perpendicular to both and therefore perpendicular to the desired plane
    I get for the normal vector:
    N = -12 i + 47 j + 14 k
    now we know the point (8,3,5) lies in the plane so a line from Point P in the plane perpendicular to N lies in the plane so the dot product of that with N is 0
    (-12)(x-x1) + 47 (y -y1) +14(z-z1) = 0
    -12(x-8) + 47 (y-3) + 14 (z-5) = 0
    -12 x + 47 y + 14 z = 115
    CHECK my arithmetic!

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