The equation of the plane passing through the line of intersection of planes $2 x + y - z = 3$ and $5 x - 3 y + 4 z + 9 = 0$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5}$ is ?

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Solution

Equation of the planes are $2 x + y - z = 3$ and $5 x - 3 y + 4 z + 9 = 0$ .
The equation of the plane passing through the line of intersection of these planes is
$(2 x + y - z - 3) + λ (5 x - 3 y + 4 z + 9) = 0$
$x (2 + 5 λ) + y (1 - 3 λ) + z (4 λ - 1) + 9 λ - 3 = 0$ .......(1)
The plane is parallel to the line :
$\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5}$

Therefore,
$2 (2 + 5 λ) + 4 (1 - 3 λ) + 5 (4 λ - 1) = 0$
$18 λ + 3 = 0$
$λ = \frac{- 1}{6}$
Putting this value in equation 1, we get,
$x (2 - \frac{5}{6}) + y (1 + \frac{3}{6}) + z (- \frac{4}{6} - 1) - \frac{9}{6} - 3 = 0$

$7 x + 9 y - 10 z - 27 = 0$
This is the equation of the required plane.

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The equation of the plane passing through the line of intersection of planes 2x+y−z=3 and 5x−3y+4z+9=0 and parallel to the line x−12=y−34=z−55 is ?

The equation of the plane passing through the line of intersection of planes $2 x + y - z = 3$ and $5 x - 3 y + 4 z + 9 = 0$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5}$ is ?