Let $\text{L}_1$ and $\text{L}_2$ be the following straight line. $\text{L}_1 : \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3}$ and $\text{L}_2 : \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}$. Suppose the straight line $\text{L} : \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}$ lies in the plane containing $\text{L}_1$ and $\text{L}_2$, and passes through the point of intersection of $\text{L}_1$ and $\text{L}_2$. If the line L bisects the acute angle between the lines $\text{L}_1$ and $\text{L}_2$, then which of the following statements is/are TRUE?

Let L₁ and L₂ be the following straight line. L₁ : (x-1)/1 = y/(-1) = (z-1)/3 and L₂ : (x-1)/(-3) = y/(-1) = (z-1)/1. Suppose the straight line L : (x-α)/l = (y-1)/m = (z-γ)/(-2) lies in the plane containing L₁ and L₂, and passes through the point of intersection of L₁ and L₂. If the line L bisects the acute angle between the lines L₁ and L₂, then which of the

$L_1:\ A(1,0,1),\ \vec{d_1} = (1,-1,3)$

$L_2:\ A(1,0,1),\ \vec{d_2} = (-3,-1,1)$

$L:\ \text{through } (\alpha,1,\gamma),\ \vec{d} = (l,m,-2)$

$L \text{ bisects the acute angle between } L_1, L_2$

$\text{Check } \alpha-\gamma \text{ and } l+m$

$L_{1}:\ (1+t,\ -t,\ 1+3t)$

$L_{2}:\ (1-3s,\ -s,\ 1+s)$

$\therefore\ \text{intersection } A = (1,0,1)$

$\text{1. Bisector } \vec{d} \propto \dfrac{\vec{d_1}}{|\vec{d_1}|} + \dfrac{\vec{d_2}}{|\vec{d_2}|}$

$\text{2. Compare with } (l,m,-2) \Rightarrow l,m$

$\text{3. Use } A=(1,0,1) \text{ on } L \Rightarrow \alpha,\gamma$

$|\vec{d_1}| = |\vec{d_2}| = \sqrt{11},\qquad \vec{d_1}\cdot\vec{d_2} = 1 > 0$

$\vec{d_1} + \vec{d_2} = (-2,-2,4) \propto (-1,-1,2)$

$(l,m,-2) \parallel (-1,-1,2)$

$\Rightarrow l = 1,\quad m = 1$

$l + m = 2\ \checkmark$

$\dfrac{1-\alpha}{1} = \dfrac{0-1}{1} = \dfrac{1-\gamma}{-2}$

$\alpha = 2,\qquad \gamma = -1$

$\alpha - \gamma = 3\ \checkmark$

$\vec{d} = \vec{d_1} + \vec{d_2} \Rightarrow L \text{ lies in the plane of } L_1, L_2$

$\boxed{\text{Answer: (1) and (2)}}$