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(B) Let the line $L_1$ have direction ratios $(1, \alpha, \beta)$,$L_2$ have $(-1, 2, 1)$,and $L_3$ have $(\alpha, 1, \beta)$.Since $L_1 \perp L_2$,th

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$(1, -1)$

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(B) Let the line $L_1$ have direction ratios $(1, \alpha, \beta)$,$L_2$ have $(-1, 2, 1)$,and $L_3$ have $(\alpha, 1, \beta)$.Since $L_1 \perp L_2$,the dot product of their direction ratios is zero:$1(-1) + \alpha(2) + \beta(1) = 0 \Rightarrow -1 + 2\alpha + \beta = 0 \Rightarrow 2\alpha + \beta = 1$ (Equation $1$).Since $L_1 \parallel L_3$,their direction ratios are proportional:$\frac{1}{\alpha} = \frac{\alpha}{1} = \frac{\beta}{\beta}$.From $\frac{1}{\alpha} = \frac{\alpha}{1}$,we get $\alpha^2 = 1$,so $\alpha = 1$ or $\alpha = -1$.If $\alpha = 1$,then from Equation $1$: $2(1) + \beta = 1 \Rightarrow \beta = -1$.If $\alpha = -1$,then from Equation $1$: $2(-1) + \beta = 1 \Rightarrow \beta = 3$.However,the condition $\frac{\beta}{\beta} = 1$ must hold for $\beta 
eq 0$. If $\beta = 0$,the ratio is undefined. Checking $\alpha = 1, \beta = -1$: the ratios are $(1, 1, -1)$ and $(1, 1, -1)$,which are parallel. Thus,$(\alpha, \beta) = (1, -1)$.

If the line with direction ratios $(1, \alpha, \beta)$ is perpendicular to the line with direction ratios $(-1, 2, 1)$ and parallel to the line with direction ratios $(\alpha, 1, \beta)$,then $(\alpha, \beta)$ is

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