Step 1:

The centre of the hyperbola is $(2,1)$ and one of the asymptotes is $3x-y-5=0$.Since in an rectangular hyperbola,the asymptotes are $\perp$ to each other,the other asymptote is of the form $x+3y+k=0$.

It passes through the centre $(2,1)$

Therefore $2+3+k=0$

$\Rightarrow 5+k=0$

$\Rightarrow k=-5$

The second asymptote is $x+3y-5=0$

Step 2:

The combined equation of the asymptotes is $(3x-y-5)(x+3y-5)=0$

The equation of the rectangular hyperbola differs only in the constant term.Its equation is of the form $(3x-y-5)(x+3y-5)=k$

Step 3:

The point $(1,-1)$ lies on the rectangular hyperbola.

Therefore $(3+1-5)(1-3-5)=k$

$(-1)(-7)=k$

$\Rightarrow k=7$

The equation of the hyperbola is $(3x-y-5)(x+3y-5)=7$

On simplifying we get,

$3x^2+8xy-3y^2-20x-10y+18=0$