and $L_2: \frac{}{3} = \frac{}{2} = \frac{}{\gamma}$ 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 the statement in Column-I with the values in Column-II 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) $\sqrt{\frac{2}{3}}$ |

and L₂: /3 = /2 = /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 the statement in Column-I with the values in Column-II 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 possible choice for n̂ is | (2) √3/2 | | (R)

$\text{Given: } L_1: \dfrac{x+1}{1} = \dfrac{y+1}{2} = \dfrac{z-1}{3}$

$L_2: \dfrac{x-2}{3} = \dfrac{y-1}{2} = \dfrac{z-2}{\gamma}$

$L_1, L_2 \text{ intersect at } R_1;\quad O = (0,0,0)$

$\text{To find: match } (P), (Q), (R), (S) \text{ with } (1)\text{--}(5)$

$\hat{n} = \text{unit normal to the plane containing } L_1, L_2$

$\vec{d_1} = (1,2,3)$

$\vec{d_2} = (3,2,\gamma)$

$L_1: (x,y,z) = (-1,-1,1) + t(1,2,3)$

$L_2: (x,y,z) = (2,1,2) + s(3,2,\gamma)$

$-1+t = 2+3s \Rightarrow t - 3s = 3$

$-1+2t = 1+2s \Rightarrow t - s = 1$

$1+3t = 2+\gamma s \Rightarrow 3t - \gamma s = 1$

$t - s = 1 \Rightarrow t = s + 1$

$(s+1) - 3s = 3 \Rightarrow s = -1,\quad t = 0$

$3(0) - \gamma(-1) = 1 \Rightarrow \gamma = 1 \Rightarrow (P) \to (3)$

$R_1 = (-1+t,\ -1+2t,\ 1+3t) = (-1,-1,1)$

$\overrightarrow{OR_1} = -\hat{i} - \hat{j} + \hat{k} \Rightarrow (R) \to (1)$

$\hat{n} \parallel \vec{d_1} \times \vec{d_2}$

d₁⃗ × d₂⃗ = î ĵ k̂ 1 2 3 3 2 1

$= \hat{i}(2-6) - \hat{j}(1-9) + \hat{k}(2-6)$

$= -4\hat{i} + 8\hat{j} - 4\hat{k} = -4(\hat{i} - 2\hat{j} + \hat{k})$

$|\vec{d_1} \times \vec{d_2}| = \sqrt{16+64+16} = 4\sqrt{6}$

$\hat{n} = \dfrac{1}{\sqrt{6}}\hat{i} - \dfrac{2}{\sqrt{6}}\hat{j} + \dfrac{1}{\sqrt{6}}\hat{k} \Rightarrow (Q) \to (4)$

$\overrightarrow{OR_1} \cdot \hat{n} = \dfrac{(-1)(1)+(-1)(-2)+(1)(1)}{\sqrt{6}} = \dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}$

$\Rightarrow (S) \to (5)$

$(P)\to(3),\quad (Q)\to(4),\quad (R)\to(1),\quad (S)\to(5) \Rightarrow \text{Option (3)}$