# Find the intersection point of line x-2/1=y+3/2=z/-2 and plane -12x+3y+4z=9

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You need to write the parametric form of the line such that:

`x = 2 + t `

`y = -3 + 2t `

`z = -2t`

You should know that the intersection point lies on the line and in plane, hence the coordinates of intersection point need to verify the equation of the line and equation of the plane, both.

You should come up with the notation for the intersection point such that: `I(x_I,y_I,z_I)`

You need to substitute the coordinates of intersection point in equation of line and plane such that:

`x_I = 2 +` t

`y_I = -3 + 2t`

`z_I = -2t`

`12x_I + 3y_I + 4z_I = 9`

You need to substitute the parametric equations of the line in equation of the plane such that:

`12(2 + t)+ 3(-3 + 2t) + 4(-2t) = 9`

You need to open the brackets such that:

`24 + 12t - 9 + 6t - 8t = 9`

`10t = -6 =gt t = -6/10 =gt t = -3/5`

You need to find the coordinates of intersection point, hence, you shoud substitute `-3/5` for t in equations of line such that:

`x_I = 2 - 3/5 =gt x_ I = 7/5`

`y_I = -3 - 6/5 =gt y_I = -21/5`

`z_I = 6/5`

**Hence, evaluating the coordinates of intersection point between line and plane yields `I(7/5 ; -21/5 ; 6/5).` **