Verified answer

There are a couple of ways to approach this.

1) find the cross-product of the direction vector of the line and another vector between the given point and a point on the line to find the normal of the plane. Use that normal to find the equation of the plane.

2) find three points in the plane (two on the line, and the given point). Use these to write a matrix equation for the plane. Solve that equation.

Method 1)

The direction vector of the line is given by the coefficients of t: (6, -4, 0). A point on the line is given by t=0: (-3, 7, -2), so another vector in the plane is

.. (0, 3, -4) -(-3, 7, -2) = (3, -4, -2)

The cross product of these vectors* (6, -4, 0) and (3, -4, -2) is (8, 12, -12), so a suitable normal vector for the plane will be this vector divided by 4: (2, 3, -3).

The equation of the plane is

.. 2x +3y -3z = 2*0 +3*3 -3*(-4)

.. 2x +3y -3z = 21

Method 2)

Two points on the line can be found from t=0 and t=1:

.. (-3, 7, -2) and (3, 3, -2)

Then we can find the values of a, b, c in the equation for the plane

.. ax +by +cz = 1

using the augmented matrix

.. [[0, 3, -4, 1] [-3, 7, -2, 1] [3, 3, -2, 1]]

A suitable calculator** or your own row-reduction process can give the solution

.. a = 2/21, b = 1/7, c = -1/7

Multiplying the above equation by 21, this gives the equation for the plane as

.. 2x +3y -3z = 21

The equation of the plane is 2x +3y -3z = 21.

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* There is a cross-product calculator at the first source link if you need one.

** There is a reduced echelon form row reduction calculator at the second source link if you need one.