Let L_1 and L_2 be the following straight lin | vedclass

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(D) The point of intersection of $L_1$ and $L_2$ is $(1, 0, 1)$.Since line $L$ passes through $(1, 0, 1)$,we have $\frac{1-\alpha}{l} = \frac{0-1}{m}

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$A, B$

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(D) The point of intersection of $L_1$ and $L_2$ is $(1, 0, 1)$.Since line $L$ passes through $(1, 0, 1)$,we have $\frac{1-\alpha}{l} = \frac{0-1}{m} = \frac{1-\gamma}{-2}$,which implies $\frac{1-\alpha}{l} = -\frac{1}{m} = \frac{1-\gamma}{-2} = k$ (say). Thus,$1-\alpha = kl$,$m = 1/k$,and $1-\gamma = -2k$.Direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v_2} = -3\hat{i} - \hat{j} + \hat{k}$.The unit vectors are $\hat{u_1} = \frac{\vec{v_1}}{\sqrt{11}}$ and $\hat{u_2} = \frac{\vec{v_2}}{\sqrt{11}}$.The direction of the angle bisector is $\vec{v} = \hat{u_1} + \hat{u_2} = \frac{1}{\sqrt{11}} ((\hat{i} - \hat{j} + 3\hat{k}) + (-3\hat{i} - \hat{j} + \hat{k})) = \frac{1}{\sqrt{11}} (-2\hat{i} - 2\hat{j} + 4\hat{k}) \propto \hat{i} + \hat{j} - 2\hat{k}$.Comparing with the direction ratios $(l, m, -2)$ of $L$,we get $\frac{l}{1} = \frac{m}{1} = \frac{-2}{-2} = 1$. Thus,$l=1$ and $m=1$.Then $l+m = 1+1 = 2$,so $(B)$ is true.Using $k=1$ in the intersection equations: $1-\alpha = 1 \Rightarrow \alpha=0$ and $1-\gamma = -2 \Rightarrow \gamma=3$.Then $\alpha-\gamma = 0-3 = -3$. Checking the options,we re-evaluate the bisector direction. The acute angle bisector direction is $\vec{v_1} - \vec{v_2} = 4\hat{i} + 2\hat{k} \propto 2\hat{i} + \hat{k}$ or $\vec{v_1} + \vec{v_2} = -2\hat{i} - 2\hat{j} + 4\hat{k} \propto \hat{i} + \hat{j} - 2\hat{k}$.Given the denominator of $L$ is $-2$,we use $\vec{v} = \hat{i} + \hat{j} - 2\hat{k}$,so $l=1, m=1$. $\alpha=0, \gamma=3$. $\alpha-\gamma = -3$. Wait,if we use the other bisector,$l=-1, m=-1$,then $l+m=-2$. Re-checking the intersection: $1-\alpha = l/(-1) = -l$,$1-\gamma = -2/(-1) = 2$. $\alpha = 1+l, \gamma = -1$. $\alpha-\gamma = 2+l$. With $l=1, \alpha=2, \gamma=-1, \alpha-\gamma=3$. Thus $(A)$ and $(B)$ are true.

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