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Q.
The point of intersection of the line $x + 1= \frac{y + 3}{3} = \frac{-z + 2}{2}$ with the plane 3x + 4y + 5z = 10 is:
Three Dimensional Geometry
Solution:
Let $\frac{x+1}{1} = \frac{y+3}{3} = \frac{-z+2}{2} = k $
Thus any point on this line will have coordinates,
x = k - 1, y = 3k - 3, z = - 2k + 2
This line intersects the plane
3x + 4y + 5z = 10. Hence value of x, y, z must satisfy equation of plane.
$\Rightarrow $ 3(k - 1) + 4(3k - 3) + 5 (-2k + 2) = 10
$\Rightarrow $ 3k - 3 + 12k - 12 + 10 - 10k = 10
$\Rightarrow $ 5k - 5 = 10
$\Rightarrow $ k = 3
$\therefore $ The point of intersection is (x, y, z)
$\Rightarrow $ x = 3 - 1, y = 3 $\times$ 3 - 3, z = - 2 $\times$ 3 + 2
$\Rightarrow $ (x, y, z) = (2, 6 - 4)