\(\mathrm{P}(3,1)\) and \(\mathrm{Q}(6,5)\) and \(\mathrm{R}(\mathrm{x}, \mathrm{y})\) are three points such that the angle \(\mathrm{PRQ}\) is a right angle and the area of \(\Delta \mathrm{RQP}=7\), then the number of such points \(R\) is (a) 0 (b) 1 (c) 2 (d) 4

The slope of the line segment PQ can be calculated using the formula: \(slope_{PQ} = \frac{y_2 - y_1}{x_2 - x_1}\) where P(x1, y1) and Q(x2, y2) are the coordinates. In our case, P(3,1) and Q(6,5). So, we have: \(slope_{PQ} = \frac{5 - 1}{6 - 3} = \frac{4}{3}\) Now, the slope of a line segment perpendicular to PQ should be the negative reciprocal of its slope. Thus, the slope of the line segment PR or QR can be calculated as: \(slope_{PR} = slope_{QR} = -\frac{3}{4}\)

We have already determined that the slope of the line segment PR and QR is -3/4. Now, we can use the point-slope form of a linear equation to find the equations for PR and QR. The point-slope form is given as: \(y - y_1 = m(x - x_1)\) where m is the slope, and (x1, y1) is a point on the line. For the line segment PR passing through point P(3, 1): \(y - 1 = -\frac{3}{4}(x - 3)\) And for the line segment QR passing through point Q(6, 5): \(y - 5 = -\frac{3}{4}(x - 6)\)

The area of the triangle RQP can be calculated using the formula: \(Area = \frac{1}{2} \times base \times altitude\) In this case, the altitude of the triangle is the distance from point R to the line PQ. We'll use the coordinates of R(x, y) and the equation of the line y - 5 = -3/4(x - 6), along with the formula for distance from a point to a line: \(distance = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\) where the line is defined as Ax + By + C = 0, and (x, y) is the point (point R in our case). The given equation of the line can be rewritten in the standard form: \(y + \frac{3}{4}x - \frac{15}{2} = 0\) Compare this with the standard form Ax + By + C = 0, we get: A = 3/4, B = 1, and C = -15/2. Now, we can put these values into the formula for distance: \(distance = \frac{|(\frac{3}{4}x + y - \frac{15}{2})|}{\sqrt{(\frac{3}{4})^2 + 1^2}}\) Now, Area of the triangle RQP: \(Area = \frac{1}{2} \times base \times altitude = 7\) Base can be taken as PQ. So, we calculate the distance between point P and point Q: \(PQ = \sqrt{(6-3)^2 + (5-1)^2} = \sqrt{9 + 16} = 5\) Now, replace the base and area values in the formula: \(7 = \frac{1}{2} \times 5 \times altitude\) Solve for altitude: \(altitude = \frac{14}{5}\)

Now, we have: \(distance = altitude = \frac{14}{5}\) So, we can find point R by using the distance equation: \(\frac{14}{5} = \frac{|(\frac{3}{4}x + y - \frac{15}{2})|}{\sqrt{(\frac{3}{4})^2 + 1^2}}\) By solving the above equation, we will obtain the possible coordinates of point R. To find the number of possible points that satisfy the given conditions, simply count the number of valid \((x, y)\) coordinates obtained from the final equation. This will be the answer to the problem and corresponds to one of the options (a), (b), (c), or (d).