Find the equation for the plane through the points Upper P0 (-2 ,2 ,-5),Q0 (1,2,-1), and Upper R0 (-1,-5,4 ). The equation of plane is:

28x - 23y - 21z = 3  

Step-by-step explanation:

First, we need to find two vectors in the plane as:

vector PQ = Q - P = (1, 2, -1) - (-2, 2, -5) = (3, 0, 4)

vector PR = R - P = (-1, -5, 4) - (-2 ,2 ,-5) = (1, -7, 9)

Then, we need to find a normal vector to the plane as:

PQ x RQ = ((0*9)-(4*-7), -(3*(9)-(4*1), (3*-7)-(0*1))

PQ x RQ = (28, -23, -21)

Finally, the equation of a plane is:

A(x-x0) + B(y-y0) + C(z-z0) = 0

Where (A,B,C) is a normal vector to the plane and (x0, y0, z0) is a point in the plane. So, replacing (A,B,C) by (28, -23, -21) and (x0, y0, z0) by P0(-2,2,-5), we can write the equation of the plane as:

28(x+2) - 23(y-2) - 21(z+5) = 0

Solving, we get:

28x + 56 - 23y + 46 - 21z - 105 = 0

28x - 23y - 21z - 3 = 0

28x - 23y - 21z = 3  

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