# Point of interception.Find the interception point of the lines 2x+2y+z=9, 3x+2y+z=4, and x+y+4z=6 .

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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).