To find the equation of the plane passing through the three points P(5, 0, 1), Q(0, 3, -1), and R(-1, 2, 0), we can follow these steps:
1) First, we calculate the vectors PQ and PR by subtracting coordinates of P from Q and R respectively.
Vector PQ = Q - P = (0 - 5, 3 - 0, -1 - 1) = (-5, 3, -2)
Vector PR = R - P = (-1 - 5, 2 - 0, 0 - 1) = (-6, 2, -1)
2) We then find a normal (perpendicular) vector to the plane by taking the cross product of vectors PQ and PR. The cross product is a vector operation that takes two vectors and returns another vector perpendicular to both of the original vectors.
Normal Vector = PQ x PR = (1, 7, 8)
This implies that the direction ratios of the line perpendicular to the plane are 1, 7, and 8.
3) Next, we put these values in the general formula for a plane, which is:
a(x - x1) + b(y - y1) + c(z - z1) = 0
Here, (x1, y1, z1) is any point on the plane, and a, b, and c are the direction ratios of the normal to the plane. As the plane passes through point P(5, 0, 1), we can use this as our point, and the normal vector as our direction ratios.
Therefore, the equation of the plane becomes:
1(x - 5) + 7(y - 0) + 8(z - 1) = 0
Simplifying, we have:
x + 7y + 8z - 13 = 0.
So, the equation of the plane passing through points P, Q, and R is x + 7y + 8z - 13 = 0. Therefore, option (c) 3x + y - 2z = 1 and option (d) -x + 4y - 3z = 2 are not correct. Checking the remaining options, we see that neither (a) 2x + 3y + z = 5 nor (b) x - 2y + 3z = 0 matches our equation.