For real numbers alpha and beta neq 0,if the | vedclass

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(C) Let the first line be $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3} = \phi$. Then any point on this line is $(\phi+\alpha, 2\phi+1, 3\phi+1)$.Le

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$7$

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(C) Let the first line be $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3} = \phi$. Then any point on this line is $(\phi+\alpha, 2\phi+1, 3\phi+1)$.Let the second line be $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3} = q$. Then any point on this line is $(q\beta+4, 3q+6, 3q+7)$.For the lines to intersect,there must exist $\phi$ and $q$ such that:$\phi+\alpha = q\beta+4$ $(i)$$2\phi+1 = 3q+6$ (ii)$3\phi+1 = 3q+7$ (iii)Subtracting (ii) from (iii),we get $\phi = 1$. Substituting $\phi=1$ into (ii),we get $2(1)+1 = 3q+6$,which implies $3q = -3$,so $q = -1$.Substituting $\phi=1$ and $q=-1$ into $(i)$,we get $1+\alpha = -\beta+4$,which simplifies to $\alpha+\beta = 3$.The point of intersection is $(\phi+\alpha, 2\phi+1, 3\phi+1) = (1+\alpha, 3, 4)$.Since this point lies on the plane $x+2y-z=8$,we have $(1+\alpha) + 2(3) - 4 = 8$.$1+\alpha+6-4 = 8 \implies \alpha+3 = 8 \implies \alpha = 5$.Since $\alpha+\beta = 3$,we have $5+\beta = 3 \implies \beta = -2$.Thus,$\alpha-\beta = 5 - (-2) = 7$.

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:

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