Determine the distance of the point from the given plane: Q = (1, 2, 3) P = (2, 1, −1) + u(1, 1, 1) + v(−1, 1, 0)

To determine the distance between a point Q and a plane defined by a point P and two direction vectors u and v, we can use the formula for the distance between a point and a plane. The formula is: Distance = |(Q - P) · n| / ||n|| where: - Q is the coordinates of the point (1, 2, 3) - P is a point on the plane (2, 1, -1) - u and v are direction vectors of the plane (1, 1, 1) and (-1, 1, 0), respectively - · represents the dot product - ||n|| represents the magnitude of the normal vector n of the plane First, we need to find the normal vector n of the plane by taking the cross product of the direction vectors u and v: n = u x v Calculating the cross product: n = (1, 1, 1) x (-1, 1, 0) n = ((1 * 1) - (1 * 0), (-1 * 1) - (1 * -1), (1 * 1) - (1 * -1)) n = (1, -1, 2) Now, we can substitute the given values into the distance formula: Distance = |(Q - P) · n| / ||n|| Q - P = (1, 2, 3) - (2, 1, -1) = (-1, 1, 4) Taking the dot product of (-1, 1, 4) and (1, -1, 2): (Q - P) · n = (-1, 1, 4) · (1, -1, 2) = (-1 * 1) + (1 * -1) + (4 * 2) = -1 - 1 + 8 = 6 Calculating the magnitude of the normal vector n: ||n|| = sqrt(1^2 + (-1)^2 + 2^2) = sqrt(1 + 1 + 4) = sqrt(6) Substituting the values into the distance formula: Distance = |(Q - P) · n| / ||n|| = |6| / sqrt(6) = 6 / sqrt(6) To simplify the expression, we can rationalize the denominator by multiplying the numerator and denominator by sqrt(6): Distance = (6 / sqrt(6)) * (sqrt(6) / sqrt(6)) = 6sqrt(6) / 6 = sqrt(6) Therefore, the distance between the point Q(1, 2, 3) and the plane defined by P(2, 1, -1), u(1, 1, 1), and v(-1, 1, 0) is $$ \sqrt{6} $$.