The equation of the plane passing through the line of intersection of the planes 2 x − y + z − 3 = 0 and 4 x − 3 y + 5 z + 9 = 0 and parallel to the line 2 x + 1 = 4 y + 3 = 5 z − 3 is αx + β y

Question

The equation of the plane passing through the line of intersection of the planes 2 x − y + z − 3 = 0 and 4 x − 3 y + 5 z + 9 = 0 and parallel to the line 2 x + 1 ​ = 4 y + 3 ​ = 5 z − 3 ​ is αx + β y + γ z + d = 0 (where α , β , γ , d are coprime integers and α > 0 ). Then α + β + γ + d =

TestHub · Accepted Answer

The equation of any plane passing through the line of intersection of the given planes is:

$(2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$

$(2 + 4 λ) x + (- 1 - 3 λ) y + (1 + 5 λ) z + (- 3 + 9 λ) = 0$

Since this plane is parallel to the line $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}$ , the normal to the plane is perpendicular to the line. Therefore, the dot product of their direction ratios is zero:

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

$4 + 8 λ - 4 - 12 λ + 5 + 25 λ = 0$

$21 λ + 5 = 0 \Rightarrow λ = - \frac{5}{21}$

Substituting $λ = - \frac{5}{21}$ into the equation of the plane, we get:

$(2 x - y + z - 3) - \frac{5}{21} (4 x - 3 y + 5 z + 9) = 0$

$21 (2 x - y + z - 3) - 5 (4 x - 3 y + 5 z + 9) = 0$

$42 x - 21 y + 21 z - 63 - 20 x + 15 y - 25 z - 45 = 0$

$22 x - 6 y - 4 z - 108 = 0$

Dividing by $2$ to make the coefficients coprime integers:

$11 x - 3 y - 2 z - 54 = 0$

Comparing this with $αx + β y + γ z + d = 0$ , we have:

$α = 11$ , $β = - 3$ , $γ = - 2$ , $d = - 54$

These are coprime integers and $α > 0$ .

$α + β + γ + d = 11 - 3 - 2 - 54 = - 48$

Answer: $- 48$

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