# Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.

The equation of a line passing through two points A(x1,y1,z1) and B(x2,y2,z2) is Given the line passes through the points A(3, –4, –5) and

B(2, –3, 1)

x1 = 3, y1 = -4, z1 = -5

and, x2 = 2, y2 = -3, z2 = 1

So, the equation of line is  So,

x = -k + 3 | y = k - 4 | z = 6k - 5 …(1)

Let (x, y, z) be the coordinates of the point where the line crosses the plane 2x + y + z + 7 = 0

Putting the value of x,y,z from (1) in the equation of plane,

2x + y + z + 7 = 0

2(-k + 3) + (k - 4) + (6k - 5) = 7

5k - 3 = 7

5k = 10

k = 2

Putting the value of k in x, y, z

x = - k + 3 = - 2 + 3 = 1

y = k - 4 = 2 - 4 = - 2

z = 6k - 5 = 12 - 5 = 7

Hence, the coordinates of the required point are (1, -2,7).

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Find the coordinate of the point P where the line through and crosses the plane passing through three points and Also, find the ratio in which P divides the line segment AB.

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