Given the system of equations $2x+9y+5z=8, 2x+3y-z=-4$ and $x-2z=-5$ have an infinite number of solutions given by$x=-5+at, y=2+bt, z=ct, t\in R$ then $x, y, z$ will satisfy the given equations.

Thus, we have $2\left(-5+at\right)+9\left(2+bt\right)+5\left(ct\right)=8, 2\left(-5+at\right)+3\left(2+bt\right)-\left(ct\right)=-4$ and $\left(-5+at\right)-2\left(ct\right)=-5$

$\Rightarrow -10+2at+18+9bt+5ct=8, -10+2at+6+3bt-ct=-4$ and $-5+at-2ct=-5$

$\Rightarrow 2at+9bt+5ct=0, 2at+3bt-ct=0$ and $at-2ct=0$

$\Rightarrow 2a+9b+5c=0, 2a+3b-c=0$ and $a-2c=0.$

From the third equation, we get $a=2c,$ put this value in any of the first two equations, to get

$2\times 2c+9b+5c=0$

$\Rightarrow b=-c.$

Thus, we have $a:b:c=2c:-c:c$

$\Rightarrow a:b:c=2:-1:1.$

Hence, the solution can be $a=2, b=-1, c=1.$