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#