# Determine a direction vector for the line of intersection of the plane through the points A(1,2,4), B(-1,2,-3) and C(3,5,-1) and the plane [x,y,z]=[3,6,7] + s[4,-5,1] + t[3,-1,-1]. The equation of a plane passing through the points A(1,2,4), B(-1,2,-3) and C(3,5,-1) is determined by finding the vectors `vec(AB)` and `vec(AC)`

`vec(AB)` = (-1-1)i + (2 - 2)j + (-3 - 4)k = -2i -7k

`vec (AC)` = (3 - 1)i + (5 - 2)j + (-1-4)k = 2i + 3j - 5k

The normal vector n = `vec (AB)xx vec(AC)` = i(21) - j(10 + 14) + k(-6) = 21i - 24j - 6k

The equation of the plane in normal form is 21(x - 1) - 24(y - 2) - 6(z - 4) = 0

The equation of the plane can be converted to normal form as follows:[x,y,z]=[3,6,7] + s[4,-5,1] + t[3,-1,-1]

`<4, -5, 1>xx<3, -1, -1>` = 6i + 7j + 11k

The equation of the plane is 6(x - 3) + 7(y - 6) + 11(z - 7) = 0

The cross product of 21i - 24j - 6k and 6i + 7j + 11k is 222i + 267j - 291k.

The direction vector of the line of intersection of the given planes is 222i + 267j - 291k.

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