$$2 = \frac{8}{R1[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$
$$3 = \frac{8}{R2[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$
$$1 = \frac{8}{R3[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$
Let $\frac{1}{R1}=x$, $\frac{1}{R2}=y$, $\frac{1}{R3}=z$
So now,
$x+y+z=4x$, that is, $3x-y-z=0 -(1)$
$x+y+z=\frac{8}{3}y$, that is, $3x-5y+3z=0 - (2)$
$x+y+z=8z$, that is, $x+y-7z=0 - (3)$
Now solve the 3 equations to get unique solution and replace $(R1, R2, R3)$ with $(\frac{1}{x},\frac{1}{y},\frac{1}{z})$ to get the required answer.
Also, the trivial solution (x,y,z)=(0,0,0) won't work for this.