We have $A(3,5), B(-1, -2)$ and $d:7x-6y+1=0$, and we have to find a point $C$ on $d$ such that area of triangle $ABC$ to be $1$.
I wrote an relation between $C's$ coordinates from $C$ belongs to $d$, and from area value I found height from $C$, $CC'$, and from that value i tried to get another relation between $C$ 's coordinates but it get complicated.
Is there another approach to solve that?
You can calculate the area using a determinant. As $x=\frac{6y-1}{7}$, you must only solve the determinant
$$\frac{1}{2}\left| det \left[ {\begin{array}{cc} 3 & 5 & 1 \\ -1 & -2 & 1 \\ x & \frac{6y-1}{7} & 1 \end{array} } \right] \right| =1 \]. $$
Solving that, you will find the $x$ and, with this valeue, it is easy fo find the $y$.