Find the intersection of the line $x - 2 y + 4 z + 4 = 0$ , $x + y + z - 8 = 0$ with the plane $x - y + 2 z + 1 = 0$ .

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Solution

Line: $x - 2 y + 4 z + 4 = 0 = x + y + z - 8$
Plane: $x - y + 2 z + 1 = 0$
The line is actually line of intersection of two plane with directions of normal $(1, - 2, 4)$ & $(1, 1, 1)$ . $(l m n)$ be directions of line.
$l - 2 m + 4 n = 0$
$l + m + n = 0$
By cramer's rule
$\frac{l}{- 6} = \frac{- m}{- 3} = \frac{n}{3} ∴ (l, m, n) = (- 2, 1, 1)$
One point on line $(x, y, z) = (0, 6, 2)$
$∴ \frac{x}{- 2} = \frac{y - 6}{1} = \frac{z - 2}{1} = k$ is eqution of line.
Let $(- 2 r, r + 6, r + 2)$ be the point on line that intersects the plane
$∴ - 2 r - r - 6 + 2 r + 4 + 1 = 0 \Rightarrow r = - 1$
$∴$ Point of intersection $= (2, 5, 1)$

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Find the intersection of the line x−2y+4z+4=0, x+y+z−8=0 with the plane x−y+2z+1=0.

Find the intersection of the line $x - 2 y + 4 z + 4 = 0$ , $x + y + z - 8 = 0$ with the plane $x - y + 2 z + 1 = 0$ .