Answer:
13x +4y -z = -9
Explanation:
You want the equation of the plane through points P(-2, 3, -5), Q(0, -3, -3), and R(1, -5, 2).
Direction
The direction vector perpendicular to the plane will be the cross product of the direction vectors of two lines in the plane:
PQ × PR = (-26, -8, 2)
Equation
We can remove a factor of -2 to get the direction vector (13, 4, -1). These values are the coefficients in the plane equation:
13x +4y -z = c . . . . . where c is the dot-product of (13, 4, -1) with any of the given points.
Using point P, we have ...
13(-2) +4(3) -(-5) = c = -26 +12 +5 = -9
The equation of the plane is 13x +4y -z = -9.
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