Let L_1: frac{x-1}{1}=frac{y-2}{-1}=frac{z-1} | vedclass

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(C) The direction ratios of $L_1$ are $\vec{v_1} = (1, -1, 2)$ and for $L_2$ are $\vec{v_2} = (-1, 2, 1)$.The direction ratios of $L_3$ are given by $

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$25$

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(C) The direction ratios of $L_1$ are $\vec{v_1} = (1, -1, 2)$ and for $L_2$ are $\vec{v_2} = (-1, 2, 1)$.The direction ratios of $L_3$ are given by $\vec{v_3} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -1 & 2 & 1 \end{vmatrix} = \hat{i}(-1-4) - \hat{j}(1+2) + \hat{k}(2-1) = -5\hat{i} - 3\hat{j} + \hat{k}$.Since $L_3$ passes through $(\alpha, \beta, \gamma)$ and intersects $L_1$,the point $(\alpha, \beta, \gamma)$ must lie on $L_1$.Thus,$\frac{\alpha-1}{1} = \frac{\beta-2}{-1} = \frac{\gamma-1}{2} = k$.So,$\alpha = k+1$,$\beta = -k+2$,and $\gamma = 2k+1$.Substituting these into the expression $|5\alpha - 11\beta - 8\gamma|$:$|5(k+1) - 11(-k+2) - 8(2k+1)| = |5k + 5 + 11k - 22 - 16k - 8| = |(5+11-16)k + (5-22-8)| = |-25| = 25$.

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