How Do You Find the Equation of a Plane Through a Point and Containing a Line?

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To find the equation of a plane through the point (1,3,1) and containing the line defined by x=t, y=t, z=-2+t, first identify the directional vector of the line, which is (1, 1, 1). Next, select a point on the line, such as (0, 0, -2), and calculate the vector from this point to (1, 3, 1), resulting in the vector (1, 3, 3). With the two directional vectors, the equation of the plane can be expressed as P = (x,y,z) + sd1 + td2, where s and t are scalar multiples. There are multiple valid equations for the plane based on the chosen directional vectors.
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how do i work this?

find an equation for the plane that passes through the point (1,3,1) and contains the line x=t,y=t,z=-2+t

would the line equation be -2i+t(i+j+k)? and then would that mean that the direction or normal vector for the plane be i+j+k? and then what do you do? i have the answer and i can't get it...
 
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to find the equation of a plane you need to have TWO distinct directional vectors and one point the plane passes through

you have been told that it contains the line - what is the directional vector of that line
you need one more directional vector - a vector between the line and the point will give you this. Pick a point on the given line and find the vector between this point and the given point.
the equation of the plane will look like this

P = (x,y,z) + sd_{1} + td_{2}
where s and t are scalar mutliples, d1 and d2 the directional vectors and x y z is the point

p.s. there are many possible answers
 
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