# Find the coordinate of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane, passing through the points (2, 2, 1), (3, 0, 1) and (4, –1, 0).

The equation of the straight line passing through (3, –4, –5) and (2, –3, 1) is  The coordinates of any point on above line is

(-α+3, α-4, 6α-5).

Equation of the plane passing through (2, 2, 1), (3, 0, 1) and (4, –1, 0) (x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0

2x+y+z-7=0

It is given that (-α+3, α-4, 6α-5) point lies on above plane so,

2(-α+3) +α-4+6α-5-7=0

5α-10=0

α=2

Putting α = 2 in (-α+3, α-4, 6α-5), we get

(-2+3, 2-4, 6× 2-5) = (1, -2, 7)

Hence, (1, -2, 7) is the required coordinate that where the line through (3, –4, –5) and (2, –3, 1) crosses the plane, passing through the points (2, 2, 1), (3, 0, 1) and (4, –1, 0).

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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.

Mathematics - Board Papers