Answer :
The scalar equation of the plane that contains given points and lines can be written as: (-4)(x-1) + 4(y-2) + (-10)(z-3) = 0.
What is equation of plane?
A given point p(x, y, z) that is located on the plane and a normal vector n define the equation of a plane. A plane's vector and scalar components can be used to formulate the equation for the plane.
The parametric form of line equation,
[tex]p = p_0 + t. \vec v[/tex]
On comparing with the given equation in question, we can say that,
p = {x, y, z}, p₀ = {0, 6, 1,}, vec(v) = {3, -2, -2}
The scalar equation can be given by,
[tex][ \vec \omega, p - p_1] = 0[/tex]
Here, p₁ = {1, 2, 3} and ω is perpendicular to v and to the segment p₁ - p₀ So,
[tex]\vec \omega = \vec v \times (p_1 - p_0) = ( 3, -2, -2 ) \times (1, -4, -2)[/tex]
then,
[tex]\vec \omega = (-4, 4, -10)[/tex]
Therefore, the scalar equation can be written as:
(-4)(x-1) + 4(y-2) + (-10)(z-3) = 0
It can also be written as:
26 - 4x + 4y - 10z = 0
To know more about equation of line, go to link
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Complete Question
Find a scalar equation of the plane that contains the point (1,2,3) and contains the line represented by the vector equation r(t) = [3t, 6 - 2t, 1 -2t] ?