Final answer:
To determine the equation of the plane passing through the points (3, 1, -1), (4, 0, 2), and (6, 3, 1), find two vectors in the plane and then find the cross product of these vectors to obtain the normal vector. Finally, use the coordinates of one of the points and the normal vector to write the equation of the plane in scalar form.
Step-by-step explanation:
To determine the equation of the plane passing through the points (3, 1, -1), (4, 0, 2), and (6, 3, 1), first find two vectors in the plane. We can find two vectors by subtracting the coordinates of one point from the coordinates of the other two points:
Vector v1 = (4, 0, 2) - (3, 1, -1) = (1, -1, 3)
Vector v2 = (6, 3, 1) - (3, 1, -1) = (3, 2, 2)
Next, find the cross product of the two vectors to find the normal vector of the plane. The cross product is found by taking the determinants of the following matrix:
|i j k|
|1 -1 3|
|3 2 2|
Evaluating the determinants, the cross product is (8, -7, -5). Therefore, the equation of the plane passing through the points is 8(x-3) - 7(y-1) - 5(z+1) = 0.