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 $x+2y-z=8$, then $\alpha-\beta$ is equal to :
Show Hint
To find the intersection of two lines in 3D, parametrize each line with a different parameter ($\lambda$ and $\mu$). Equate the corresponding x, y, and z coordinates to get a system of equations. Solve for $\lambda$ and $\mu$ using two of the equations, then check for consistency with the third.
Let the two lines be L1 and L2.
L1: $\frac{x-\alpha}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda$ (say)
Any point on L1 can be written as $(\lambda+\alpha, 2\lambda+1, 3\lambda+1)$.
L2: $\frac{x-4}{\beta} = \frac{y-6}{3} = \frac{z-7}{3} = \mu$ (say)
Any point on L2 can be written as $(\beta\mu+4, 3\mu+6, 3\mu+7)$.
Since the lines intersect, there is a common point for some values of $\lambda$ and $\mu$.
Equating the coordinates:
1) $\lambda+\alpha = \beta\mu+4$
2) $2\lambda+1 = 3\mu+6 \implies 2\lambda - 3\mu = 5$
3) $3\lambda+1 = 3\mu+7 \implies 3\lambda - 3\mu = 6 \implies \lambda - \mu = 2$
From equation (3), $\lambda = \mu+2$.
Substitute this into equation (2):
$2(\mu+2) - 3\mu = 5 \implies 2\mu+4-3\mu=5 \implies -\mu = 1 \implies \mu = -1$.
Then $\lambda = -1+2 = 1$.
Now we have the parameters for the point of intersection. Let's find the coordinates of this point using L1 and $\lambda=1$.
Intersection point P = $(1+\alpha, 2(1)+1, 3(1)+1) = (1+\alpha, 3, 4)$.
This point of intersection lies on the plane $x+2y-z=8$.
Substitute the coordinates of P into the plane equation:
$(1+\alpha) + 2(3) - (4) = 8$.
$1+\alpha + 6 - 4 = 8$.
$\alpha + 3 = 8 \implies \alpha = 5$.
Now we can find $\beta$ using equation (1) with $\lambda=1, \mu=-1, \alpha=5$.
$\lambda+\alpha = \beta\mu+4$.
$1+5 = \beta(-1)+4$.
$6 = -\beta + 4$.
$\beta = 4 - 6 = -2$.
The question asks for the value of $\alpha - \beta$.
$\alpha - \beta = 5 - (-2) = 5+2 = 7$.
