Question

**Given:** Point $P$ lies on $\dfrac{x-1}{2} = \dfrac{y+1}{-3} = z = \lambda$, so $P = (2\lambda+1,\ -3\lambda-1,\ \lambda)$. **Finding $\lambda$:** Distance from $(1,-1,0)$ to $P$: $$\sqrt{(2\lambda)^2 + (-3\lambda)^2 + (\lambda)^2} = \sqrt{14\lambda^2} = |\lambda|\sqrt{14} = 4\sqrt{14}$$ $$\Rightarrow |\lambda| = 4 \Rightarrow \lambda = \pm 4$$ **Selecting the nearer point to origin:** - $\lambda = 4$: $P = (9, -13, 4)$, $|OP|^2 = 81+169+16 = 266$ - $\lambda = -4$: $P = (-7, 11, -4)$, $|OP|^2 = 49+121+16 = 186$ Since $\lambda = -4$ gives the point nearer to the origin: $$P = (-7,\ 11,\ -4) \Rightarrow (\alpha,\beta,\gamma) = (-7, 11, -4)$$ **Shortest distance between the two lines:** Line $L_1$: passes through $A = (-7, 11, -4)$ with direction $\vec{d_1} = (1, 2, 3)$ Line $L_2$: passes through $B = (-5, 10, 3)$ with direction $\vec{d_2} = (2, 1, 1)$ $$\vec{AB} = B - A = (2, -1, 7)$$ $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(1-6) + \hat{k}(1-4) = (-1, 5, -3)$$ $$|\vec{d_1} \times \vec{d_2}| = \sqrt{1 + 25 + 9} = \sqrt{35}$$ $$\vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (2)(-1) + (-1)(5) + (7)(-3) = -2 - 5 - 21 = -28$$ $$\text{Shortest Distance} = \frac{|{-28}|}{\sqrt{35}} = \frac{28}{\sqrt{35}} = \frac{28}{\sqrt{5}\cdot\sqrt{7}} = 4\sqrt{\frac{7}{5}}$$ **Answer: (B)**