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Q.
The equation of a plane containing the line
$\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$ and the point (0,7, - 7) is
AIEEEAIEEE 2012Three Dimensional Geometry
Solution:
The equation of a plane containing the line
$\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$ is a $(x + 1) + b (y - 3) + c (z + 2) = 0$ where
$- 3a + 2b + c = 0 \quad\quad...\left(A\right)$
This passes through $\left(0, 7, - 7\right)$
$\therefore a \left(0 + 1\right) + b \left(7 - 3\right) + c \left(- 7 + 2\right) = 0$
$\Rightarrow a + 4b - 5c = 0 \quad\quad...\left(B\right)$
On solving equation $\left(A\right)$ and $\left(B\right)$ we get
$a=1, b=1, c=1$
$\therefore $ Required plane is
$x + 1 + y - 3 + z + 2 = 0$
$\Rightarrow x+y + z= 0$