This question is far easier than it appears at first.
The key idea is that if a a linear equation has 1 variable, there will be one solution, but as soon as there are 2 variable, 2 equations are required and likewise for 3 variables, there must be 3 equations.
We have all of these scenarios presented here.
Solve the 3rd equation first as it only has 1 variable,
5z = 75
color(blue)(z =15)" "larr now use this value for z in the 2nd equation:
15x +2color(blue)(z) = -195
15x +2color(blue)((15)) = -195
15x +color(blue)(30) = -195
15x = -225
color(green)(x =-15)" "larr use this value for x in the first equation
7color(green)(x)+5y+7color(blue)(z)=60
7color(green)((-15))+5y+7color(blue)((15))=60
-105 +5y +105 = 60
5y=60
y=12