For real numbers $\alpha$ and $\beta \neq 0$, if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$, lies on the plane $x+2 y-z=8$, then $\alpha-\beta$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Three Dimensional Geometry

For real numbers

α

and

β \neq = 0

, if the point of intersection of the straight lines

\frac{x - α}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}

and

\frac{x - 4}{β} = \frac{y - 6}{3} = \frac{z - 7}{3}

, lies on the plane

x + 2 y - z = 8

, then

α - β

is equal to :

Answer: (C)

Parametrize the first line as

(α + t, 1 + 2 t, 1 + 3 t)

and the second as

(4 + β s, 6 + 3 s, 7 + 3 s)

. At intersection:

α + t = 4 + β s

1 + 2 t = 6 + 3 s

1 + 3 t = 7 + 3 s

. From the last two:

2 t - 3 s = 5

and

3 t - 3 s = 6 \Rightarrow t = 1, s = - 1

. Substituting:

α + 1 = 4 - β \Rightarrow α + β = 3

. The intersection point is

(α + 1, 3, 4)

. On the plane:

(α + 1) + 2 (3) - 4 = 8 \Rightarrow α + 3 = 8 \Rightarrow α = 5

. Then

β = - 2

, so

α - β = 5 - (- 2) = 7

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option