Solve this following

Question:

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:

 

  1. 5

  2. 9

  3. 3

  4. 7


Correct Option: , 4

Solution:

First line is $(\phi+\alpha, 2 \phi+1,3 \phi+1)$

and second line is $(\mathrm{q} \beta+4,3 \mathrm{q}+6,3 \mathrm{q}+7)$.

For intersection        $\phi+\alpha=\mathrm{q} \beta+4$ ............(I)

                                 $2 \phi+1=3 q+6$  .................(I)

                                  $3 \phi+1=3 q+7$ ................(III)

for (ii) \& (iii) $\phi=1, \mathrm{q}=-1$

So, from (i) $\alpha+\beta=3$

Now, point of intersection is $(\alpha+1,3,4)$

It lies on the plane.

Hence, $\alpha=5 \& \beta=-2$

 

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