Question 1204569
<font color=black size=3>
The given system
{{{system(2x+2y+5z=a,-3x+y-2z=b,x+z = c)}}}
converts to this matrix
{{{(matrix(3,4,2, 2, 5, a, -3, 1, -2, b, 1, 0, 1, c))}}}
Normally matrices do not have separating lines, but I think it's beneficial to have them. 
I'll place the items in a table like this
<table border = "1" cellpadding = "5"><tr><td>2</td><td>2</td><td>5</td><td>a</td></tr><tr><td>-3</td><td>1</td><td>-2</td><td>b</td></tr><tr><td>1</td><td>0</td><td>1</td><td>c</td></tr></table>


Let's apply row operations to zero out the entries below each pivot.

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td>(1/2)*R1 --> R1</td></tr><tr><td>-3</td><td>1</td><td>-2</td><td>b</td><td></td></tr><tr><td>1</td><td>0</td><td>1</td><td>c</td><td></td></tr></table>
Notation like (1/2)*R1 --> R1 means we take half of each element in row 1 (aka R1). Then store the results in R1.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>4</td><td>11/2</td><td>3a/2+b</td><td>R2 + 3*R1 --> R2</td></tr><tr><td>1</td><td>0</td><td>1</td><td>c</td><td></td></tr></table>
Something like R2 + 3*R1 --> R2 will mean that we triple everything in R1, then add those results to R2. Store the results in R2.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>4</td><td>11/2</td><td>3a/2+b</td><td></td></tr><tr><td>0</td><td>-1</td><td>-3/2</td><td>c  -  a/2</td><td>R3  -  R1 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>1</td><td>11/8</td><td>3a/8+b/4</td><td>(1/4)*R2 --> R2</td></tr><tr><td>0</td><td>-1</td><td>-3/2</td><td>c  -  a/2</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>1</td><td>11/8</td><td>3a/8+b/4</td><td></td></tr><tr><td>0</td><td>0</td><td>-1/8</td><td>-a/8 + b/4 + c</td><td>R3 + R2 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>1</td><td>11/8</td><td>3a/8+b/4</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>a-2b-8c</td><td>-8*R3 --> R3</td></tr></table>


At this point the matrix is in row echelon form (REF), but we haven't reached RREF just yet. 
Let's keep row reducing until all of the non-pivot entries are turned to 0.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>5/2</td><td>a/2</td><td></td></tr><tr><td>0</td><td>1</td><td>0</td><td>-a+3b+11c</td><td>R2  -  (11/8)*R3 --> R2</td></tr><tr><td>0</td><td>0</td><td>1</td><td>a-2b-8c</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>0</td><td>-2a+5b+20c</td><td>R1  -  (5/2)*R3 --> R3</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-a+3b+11c</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>a-2b-8c</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>0</td><td>-a+2b+9c</td><td>R1  -  R2 --> R1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-a+3b+11c</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>a-2b-8c</td><td></td></tr></table>

The matrix is now in Reduced Row Echelon Form (RREF).


The system has <font color=red>exactly one solution</font>, and it is when:
<font color=red size=4>x = -a+2b+9c
y = -a+3b+11c
z = a-2b-8c</font>


Confirmation using WolframAlpha
<a href="https://www.wolframalpha.com/input/?i=2x%2B2y%2B5z%3Da%2C-3x%2By-2z%3Db%2Cx%2Bz%3Dc">https://www.wolframalpha.com/input/?i=2x%2B2y%2B5z%3Da%2C-3x%2By-2z%3Db%2Cx%2Bz%3Dc</a>
The search input to type in is "<font color=blue>2x+2y+5z=a,-3x+y-2z=b,x+z=c</font>" without quotes.


The CAS feature in GeoGebra can also be used.
There are two options to input
Either 
<font color=blue>Solve[{2x+2y+5z=a,-3x+y-2z=b,x+z=c}]</font>
or
<font color=blue>ReducedRowEchelonForm[{{2, 2, 5, a}, {-3, 1, -2, b}, {1, 0, 1, c}}]</font>
The way GeoGebra handles matrices is that they are a collection of lists. 
Each list is enclosed in curly braces, which represents a particular row.
There are likely other CAS based matrix calculators out there that can offer similar features.



Some more practice with matrix row reduction
<a href="https://www.algebra.com/algebra/college/linear/Linear_Algebra.faq.question.1204100.html">https://www.algebra.com/algebra/college/linear/Linear_Algebra.faq.question.1204100.html</a>
and 
<a href="https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1203997.html">https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1203997.html</a>
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