Let $$ P(\alpha,\beta, \gamma)$$ be the point on the line $$\frac{x-1}{2}=\frac{y+1}{-3}=z$$ at a distance $$4\sqrt{14}$$ from the point (1, -1, 0) and nearer to the origin. Then the shortest di stance, between the Lines $$\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$$ and $$\frac{x+5}{2}= \frac{y-10}{1}=\frac{z-3}{1}$$, is equal to

A general point on the line is $$P(2k+1, -3k-1, k)$$. Its distance from $$(1, -1, 0)$$ is $$4\sqrt{14}$$:

$$\sqrt{(2k)^2 + (-3k)^2 + k^2} = 4\sqrt{14} \implies \sqrt{14k^2} = 4\sqrt{14} \implies k = \pm 4$$

• For $$k = 4$$, $$P_1 = (9, -13, 4)$$

• For $$k = -4$$, $$P_2 = (-7, 11, -4)$$ (closer to the origin)

Thus, $$(\alpha, \beta, \gamma) = (-7, 11, -4)$$.

• $$L_1$$: Passes through $$\vec{a}_1 = (-7, 11, -4)$$, direction $$\vec{b}_1 = (1, 2, 3)$$

• $$L_2$$: Passes through $$\vec{a}_2 = (-5, 10, 3)$$, direction $$\vec{b}_2 = (2, 1, 1)$$

$$\vec{a}_2 - \vec{a}_1 = (2, -1, 7)$$

$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = -1\hat{i} + 5\hat{j} - 3\hat{k}$$

• Magnitude $$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 25 + 9} = \sqrt{35}$$

• Dot Product $$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 2(-1) + (-1)(5) + 7(-3) = -28$$

$$\text{S.D.} = \frac{|-28|}{\sqrt{35}} = \frac{28}{\sqrt{35}} = 4\sqrt{\frac{7}{5}}$$