Let $L_1$ and $L_2$ be the following straight line. $L_1: \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3}$ and $L_2: \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}$. Suppose the straight line $L: \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}$ lies in the plane containing $L_1$ and $L_2$, and passes through the point of intersection of $L_1$ and $L_2$. If the line L bisects the acute angle between the lines $L_1$ and $L_2$, 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 following statements is/are TRUE?

$L_1:\ \dfrac{x-1}{1}=\dfrac{y}{-1}=\dfrac{z-1}{3}$

$L_2:\ \dfrac{x-1}{-3}=\dfrac{y}{-1}=\dfrac{z-1}{1}$

$L:\ \dfrac{x-\alpha}{l}=\dfrac{y-1}{m}=\dfrac{z-\gamma}{-2}$

$L \text{ lies in the plane of } L_1, L_2 \ \text{and passes through } L_1\cap L_2$

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

$\text{Find the TRUE statement(s) among (1)--(4)}$

$L_1:\ (1,0,1), \quad \vec{d}_1=(1,-1,3)$

$L_2:\ (1,0,1), \quad \vec{d}_2=(-3,-1,1)$

$L_1\cap L_2=(1,0,1)$

$|\vec{d}_1|=\sqrt{1+1+9}=\sqrt{11}$

$|\vec{d}_2|=\sqrt{9+1+1}=\sqrt{11}$

$\Rightarrow |\vec{d}_1|=|\vec{d}_2|$

$\text{Plan: } 1.\ L_1\cap L_2=(1,0,1)$

$2.\ \text{bisector} \propto \dfrac{\vec{d}_1}{|\vec{d}_1|}+\dfrac{\vec{d}_2}{|\vec{d}_2|}$

$3.\ (l,m,-2)\parallel \text{bisector};\ (1,0,1)\in L \Rightarrow \alpha,\gamma$

$\dfrac{\vec{d}_1}{|\vec{d}_1|}+\dfrac{\vec{d}_2}{|\vec{d}_2|}=\dfrac{(1,-1,3)+(-3,-1,1)}{\sqrt{11}}$

$=\dfrac{(-2,-2,4)}{\sqrt{11}}=\dfrac{2\,(-1,-1,2)}{\sqrt{11}}$

$\text{acute bisector} \propto (-1,-1,2) \propto (1,1,-2)$

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

$\Rightarrow l+m=2 \quad \text{: (2) TRUE, (4) FALSE}$

$(1,0,1)\in L:\ \dfrac{1-\alpha}{l}=\dfrac{0-1}{m}=\dfrac{1-\gamma}{-2}$

$1-\alpha=-1 \Rightarrow \alpha=2$

$\dfrac{1-\gamma}{-2}=-1 \Rightarrow 1-\gamma=2 \Rightarrow \gamma=-1$

$\alpha-\gamma=2-(-1)=3$

$\Rightarrow \text{statement (1) TRUE, (3) FALSE}$

$\vec{d}_1\times\vec{d}_2=(2,-10,-4)=2(1,-5,-2)$

$\vec{n}\cdot(l,m,-2)=(1)(1)+(-5)(1)+(-2)(-2)=0$

$\Rightarrow L\parallel \text{plane and } (1,0,1)\in\text{plane} \Rightarrow L\subset\text{plane}$

$\therefore\ \text{TRUE statements: (1) and (2)}$

$\alpha-\gamma=3 \quad \text{and}\quad l+m=2$