Question 824806
<pre>
FIRST DETERMINANT = {{{abs(matrix(3,3,
x,y,z,
a,b,c,
2,4,5))}}}{{{""=""}}}{{{3}}},  SECOND DETERMINANT = {{{abs(matrix(3,3,
2,4,5,
3a,3b,3c,
x-2,y-4,z-5))}}}{{{""=""}}}{{{"?"}}}

We notice that 

1. the second determinant's top row is the same as
the first matrix's bottom row. 
2. we notice that the second determinant's middle row is 
the first determinant's middle row multiplied by 3.
3. we notice that the second determinant's bottom row is
the first determinant's top row with the first determinant's
bottom row subtracted from it.

So let's start with the first determinant 

{{{abs(matrix(3,3,
x,y,z,
a,b,c,
2,4,5))}}}{{{""=""}}}{{{3}}}

and do row operations until we get the second determinant.  
Let's begin by swapping the top and bottom rows:

We remember the rule that when we swap two rows in a determinant,
we multiply the value of the determinant by -1, so

{{{abs(matrix(3,3,
2,4,5,
a,b,c,
x,y,z))}}}{{{""=""}}}{{{3(-1)}}}{{{""=""}}}{{{-3}}}

We notice that the second determinant's middle row is this
determinant's middle row multiplied by 3.  So we multiply
this determinant's middle row by 3.

We remember the rule that when we multiply a row in a determinant
by a scalar (constant), we multiply the value of the determinant 
by that same constant, so the value of the determinant is multiplied 
by 3:

{{{abs(matrix(3,3,
2,4,5,
3a,3b,3c,
x,y,z))}}}{{{""=""}}}{{{-3(3)}}}{{{""=""}}}{{{-9}}}

Finally we notice that the second determinant's bottom row is
this determinant's bottom row with the top row subtracted from it

We remember the rule that if we add a multiple of one row to 
another row, we do not change the value of the determinant, so
we add -1 times the top row from the bottom row and we will
not change the value of the determinant, so we have:

{{{abs(matrix(3,3,
2,4,5,
3a,3b,3c,
x-2,y-4,z-5))}}}{{{""=""}}}{{{-9}}}

Edwin</pre>