Let $\gamma \in \mathbb{R}$ be such that the lines $L_1: \frac{x+11}{1} = \frac{y+21}{2} = \frac{z+29}{3}$ and $L_2: \frac{x+16}{3} = \frac{y+11}{2} = \frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O = (0, 0, 0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$. Match each entry in List-I to the correct entry in List-II. | List-I | List-II | | --- | --- | | (P) $\gamma$ equals | (1) $-\hat{i} - \hat{j} + \hat{k}$ | | (Q) A possible choice for $\hat{n}$ is | (2) $\sqrt{\frac{3}{2}}$ | | (R) $\overrightarrow{OR_1}$ equals | (3) $1$ | | (S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is | (4) $\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$ | | | (5) $\sqrt{\frac{2}{3}}$ |

Let γ ∈ R be such that the lines L₁: (x+11)/1 = (y+21)/2 = (z+29)/3 and L₂: (x+16)/3 = (y+11)/2 = (z+4)/γ intersect. Let R₁ be the point of intersection of L₁ and L₂. Let O = (0, 0, 0), and n̂ denote a unit normal vector to the plane containing both the lines L₁ and L₂. Match each entry in List-I to the correct entry in List-II. | List-I | List-II | | --- | --- | | (P) γ equals | (1) -î - ĵ + k̂ | | (Q) A possibl $\text{Given: } L_1: \dfrac{x+11}{1} = \dfrac{y+21}{2} = \dfrac{z+29}{3}$ $L_2: \dfrac{x+16}{3} = \dfrac{y+11}{2} = \dfrac{z+4}{\gamma},\qquad \gamma \in \mathbb{R}$ $R_1 = L_1 \cap L_2,\quad O=(0,0,0),\quad \hat{n} \perp \text{plane}(L_1, L_2)$ $\text{To find: match List-I with List-II, then the correct option}$ $\text{Plan: } \gamma \text{ from intersection} \Rightarrow R_1 \Rightarrow \hat{n},\, \overrightarrow{OR_1},\, \overrightarrow{OR_1}\cdot\hat{n}$ $(P)\to?,\ (Q)\to?,\ (R)\to?,\ (S)\to?$ $L_1:\ \vec r = (-11,-21,-29) + t(1,2,3)$ $\text{Intersect} \Leftrightarrow (\vec A_2-\vec A_1)\cdot(\vec d_1\times\vec d_2) = 0$ $\vec A_2-\vec A_1 = (-16,-11,-4)-(-11,-21,-29) = (-5,10,25)$ $\vec d_1\times\vec d_2 = (2\gamma-6,\ 9-\gamma,\ -4)$ $(-5)(2\gamma-6) + 10(9-\gamma) + 25(-4) = 0$ $-20\gamma + 20 = 0 \Rightarrow \gamma = 1$ $\therefore\ (P)\to(3)$ $L_2:\ \vec r = (-16,-11,-4) + s(3,2,1)$ $L_1: (-11+t,\ -21+2t,\ -29+3t)$ $L_2: (-16+3s,\ -11+2s,\ -4+s)$ $t=10,\ s=5 \Rightarrow R_1 = (-1,-1,1)$ $(R)\ \overrightarrow{OR_1} = R_1 - O = (-1,-1,1)$ $= -\hat{i} - \hat{j} + \hat{k} \Rightarrow (R)\to(1)$ $(Q)\ \vec d_1\times\vec d_2 = (-4,8,-4),\quad |\cdot| = \sqrt{96} = 4\sqrt{6}$ $\hat n = \dfrac{(-4,8,-4)}{4\sqrt{6}} = -\dfrac{1}{\sqrt{6}}\hat i + \dfrac{2}{\sqrt{6}}\hat j - \dfrac{1}{\sqrt{6}}\hat k$ $\text{option }(4) \text{ is the negative} \Rightarrow (Q)\to(4)$ $(S)\ \overrightarrow{OR_1}\cdot\hat n = (-1,-1,1)\cdot\dfrac{1}{\sqrt6}(1,-2,1)$ $= \dfrac{-1+2+1}{\sqrt6} = \dfrac{2}{\sqrt6} = \sqrt{\dfrac{2}{3}}$ $\Rightarrow (S)\to(5)$ $(P)\to(3),\quad (Q)\to(4),\quad (R)\to(1),\quad (S)\to(5)$ $\boxed{\text{Option }(3)}$