What is the interception point of the lines 2x+2y+z=9, 3x+2y+z=4, x+y+4z=6.

To determine the intercepting point, we'll solve the
system formed by the equations of the lines.

We'll
calculate the determinant of the system. The determinant of the system is formed from
the coefficients of the variables, x, y and z.

We'll note
the determinant as det A.

             1   1 
 4

det A =  3   2  
1

              2   2  
1

We'll calculate det A:

det A
= 1*2*1 + 3*2*4  + 1*1*2 - 4*2*2 - 2*1*1 - 3*1*1

det A = 2
+ 24 + 2 - 16 - 2 - 3

We'll eliminate and combine like
terms:

det A = 7

Since the
determinant is not zero, we'll have a single solution of the system, that represents the
intercepting point of the lines.

Now, we'll calculate the
variable x using Cramer formula:

x = detX /
detA

Det X is the determinant whose column of coefficients
of the variable that has to be found (in this case x) is substituted by the column of
the terms from the right side of the equal (6 , 4 ,
9).

            6   1   4

det
X = 4   2   1

            9    2  
1

det X = 6*2*1 + 4*2*4 + 1*1*9 - 4*2*9 - 6*2*1 -
4*1*1

We'll eliminate like
terms:

detX = 32 + 9 - 72 - 12 -
4

det X = -47

x =
detX/detA

x =
-47/7

            1   6  
4

det Y = 3   4  
1

             2   9   1

det Y
= 4 + 108 + 12 - 32 - 9 - 18

y =
detY/detA

y =
65/7

We'll calculate z substituting the
values of x and y into the first equation:

x+y+ 4z =
6

4z = 6 - x - y

z = (6  -x -
y)/4

z = (6 + 47/7 - 65/7)/4

z
= -12/7*4

z =
-3/7

The coordinates of the
intercepting point are: (-47/7,65/7,-3/7).