#### Question

Find the coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)

#### Solution 1

Equation of the line passing through (3, – 4, – 5) and (2, – 3, 1) is

⇒ (x - 1)(12) + (y - 2)6 + (z - 3)6 = 0

12x - 12 + 6y - 12 + 6z -18 = 0

12x + 6y + 6z - 42 = 0

2x + y + z - 7 = 0

2(-λ + 3) + 1(λ - 4) + (6λ - 5) - 7 = 0

- 2λ + 6 + λ - 4 + 6λ - 5 -7 = 0

5λ = 10 ⇒ λ = 2

∴ x = -2 +3, y = 2 - 4, z= 12 - 5

∴ x = 1, y = -2, z = 7

∴ Intersection point is (1,-2,7)

#### Solution 2

We know that the cartesian equation of a line passing through two points (x_{1}, y_{1}, z_{1})

and (x_{2}, y_{2}, z_{2})is given by

`(x - x_1)/(x_2 - x_1) = (y - y_1)/(y_2 - y_1) = (z - z_1)/(z_2 - z_1)`

So, the equation of a line passing through (3, –4, –5) and (2, –3, 1) is

`(x-3)/(2-3) = (y-(-4))/(-3-(-4)) = (z -(-5))/(1-(-5))`

`=> (x - 3)/(-1) = (y - (-4))/1 = (z-(-5))/6`

`=> (x - 3)/(-1) = (y+4)/1 = (z+5)/6`

Now, the coordinates of any point on this line are given by

`(x-3)/(-1) = (y +4)/1 = (z +5)/6 = k`

⇒ x = 3- k,y = k - 4, z = 6k - 5, where k is a constant

Let R(3 − *k*, *k* − 4, 6*k* − 5) be the required point of intersection.

Now,

Let the equation of a plane passing through (1, 2, 3) be

a(x − 1) + b(y − 2) + c(z − 3) = 0 .....(1)

Here, *a*, *b*, *c* are the direction ratios of the normal to the plane.

Since the plane (1) passes through (4, 2, −3), so

a(4 - 1) + b(2 - 2) + c(-3 - 3) = 0

⇒ 3a - 6c = 0 .....(2)

Also, the plane (1) passes through (0, 4, 3), so *a*(0 - 1) + *b*(4 - 2) + *c*(3 - 3) = 0

⇒ -a + 2b = 0 .....(3)

Solving (2) and (3) using the method of cross multiplication, we have

`a/(0 + 12) = b/(6 - 0) = c/(6 + 0)`

`=> a/12 = b/6 = c/6`

⇒ `a/2`= b = c = λ (Say)

⇒ a = 2λ, b = λ, c = λ

From (1), we get

2λ(x−1) + λ(y−2) + λ(z−3)=0

⇒ 2x + y +z −7=0 .....(4)

Putting x = 3−k, y = k−4, z = 6k−5 in (4), we get

⇒ 2(3 - k) + (k - 4) + (6k - 5) - 7 = 0

⇒5k − 10=0

⇒k = 2

Putting *k* = 2 in R(3 − *k*, *k* − 4, 6*k* − 5), we get

R(3 - k, k - 4, 6k - 5) = R(3 - 2, 2 - 4, 6 × 2 - 5) = R(1, -2,7)

Thus, the coordinates of the required point are (1, –2, 7) .