P(3, 1), Q(6, 5) and R(x, y) are three points such that the angle RPQ is a right angle and the area of $∆$RPQ =
7, then the number of such points R is

Since the angle RPQ is a right angle,
$\therefore$ slope of RP × slope of PQ = –1

$\Rightarrow \frac{1-y}{3-x}\times \frac{5-1}{6-3}=-1\Rightarrow 3x+4y=13$               (1)

Also, area of $∆$RPQ = 7

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 3& 1& 1\\ 6& 5& 1\end{array}\right|=7$

$\Rightarrow \frac{1}{2}\left[x(1-5)-y(3-6)+1(15-6)\right]=\pm 7$

⇒ – 4x + 3y + 9 = ± 14 ⇒ – 4x + 3y = 5                       (2)
and – 4x + 3y = –23                                                      (3)
Solving Eq. (1) and (2) and (1) and (3), we get two
different coordinates of the point R. So, there are two such points R.