Question

Let the line
$\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4}$
intersect the plane containing the lines
$\frac{x - 4}{1} = \frac{y + 1}{-2} = \frac{z}{1}$ and $4ax-y+5z-7a = 0 = 2x-5y-z-3, a∈R$
at the point P(α, β, γ). Then the value of α + β + γ equals _____.

Answer 1

Equation of plane containing the line
4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3 can be written as
4ax-y+5z-7a+λ(2x-5y-z-3) = 0
(4a+2λ)x-(1+5λ)y+(5-λ)z-(7a+3λ) = 0
Which i coplanar with the line
x-4/1 = y+1/-2 = z/1
4(4a+2λ)+(1+5λ)-(7a+3λ) = 0
9a+10λ+1 = 0....(1)
(4a+2λ)1+(1+5λ)2+5-λ = 0
4a+11λ+7 = 0....(2)
a = 1, λ = -1
Equation of plane is x + 2 y + 3 z – 2 = 0
Intersection with the line
$\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4}$
(7 t + 3) + 2 (– t + 2) + 3 (– 4 t + 3) – 2 = 0
–7 t + 14 = 0
t = 2
So, the required point is (17, 0, –5)
α+β+γ = 12