# What point P do the three planes 7x+35y+132z = 1 1x+6y+23z=1 1x+5y+19z=1intersect at?

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### 1 Answer

You need to find the point of intersection of the given lines, hence, you need to solve for x,y,z the system of equations, such that:

`{(7x+35y+132z = 1),(x+6y+23z = 1),(x+5y+19z = 1):}`

Elimination of one variable is one method you may approach when evaluate the system, such that:

`x + 6y + 23z - x - 5y - 19z = 1 - 1`

`y + 4z = 0`

`7x + 35y + 132z - 7x - 42y - 161z = 1 - 7`

`-7y - 29z = -6 => 7y + 29z = 6`

You may solve the new system of equations using substitution, such that:

`{(y = -4z),(7*(-4z) + 29z = 6):} => {(y = -24),(z = 6):}`

Replacing -24 for y and 6 for z in `x+5y+19z = 1` yields:

`x = 1 - 5*(-24) - 19*6 => x = 1 + 120 - 114 => x = 7`

**Hence, evaluating the coordinates of point of intersection of the given lines, yields that they intersect at **`x = 7, y = -24, z = 6.`