Question 6156
You are looking to find one variable in terms of the others. You do this by rearranging. Rearrange the last equation to find y. (I picked that one because it looked good. Any others will work too.)
{{{9x + 2y+12z=14}}} rearranges to {{{2y=14-12z-9x}}} and then to {{{y=(14-12z-9x)/2}}}
Now substitute this value for y in the equations. The first looks like this:
{{{22x+(5(14-12z-9x)/2)+7z=12}}}
{{{22x+((70-60z-45x)/2)+7z=12}}}
{{{44x+70-60z-45x+14z=24}}}  
{{{-x-46z=-46}}}
{{{x +46z=46}}}
Do the same substitution with the second equation.
{{{10x+(3(14-12z-9x)/2)+2z=5}}}
{{{10x+((42-36z-27x)/2)+2z=5}}}
{{{20x+42-36z-27x+4z=10}}}
{{{-7x-32z=-32}}}
{{{7x+32z=32}}}
Now do the same elimination with the two equations in x and z.
The first equation says
{{{x=46-46z}}}
Substitute x in the second equation.
{{{7(46-46z)+32z=32}}}
{{{322-322z+32z=32}}}  [resist the urge to simplify!]
{{{-322z+32z=32-322}}}
{{{32z-322z=32-322}}}
{{{(32-322)z=(32-322)}}}  [factor out 32-322]
{{{z=1}}}  [divide by (32-322) !]
Now substitute the value for z in {{{x=46-46z}}}
{{{x=0}}}
Now substitute the values for x and z in {{{9x+2y+12z=14}}}
{{{2y=14-0-12}}}
{{{y=1}}}
Check in the original equations.