Question:

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 distance between the lines \[ \frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3} \quad \text{and} \quad \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1} \] is equal to

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For shortest distance between two skew lines, always use the vector cross product formula involving direction vectors and a point-to-point joining vector.
Updated On: Feb 4, 2026
  • \( 7\sqrt{\frac{5}{4}} \)
  • \( 4\sqrt{\frac{5}{7}} \)
  • \( 2\sqrt{\frac{7}{4}} \)
  • \( 4\sqrt{\frac{7}{5}} \)
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The Correct Option is D

Solution and Explanation

Step 1: Find the coordinates of point \( P(\alpha,\beta,\gamma) \).

Given the line
\[ \frac{x-1}{2} = \frac{y+1}{-3} = z = t \]
\[ x = 1 + 2t,\quad y = -1 - 3t,\quad z = t \]
Distance of \( P \) from \( (1,-1,0) \) is given as \( 4\sqrt{14} \):

\[ \sqrt{(2t)^2 + (-3t)^2 + t^2} = 4\sqrt{14} \]
\[ \sqrt{14t^2} = 4\sqrt{14} \Rightarrow |t| = 4 \]
Since the point is nearer to the origin, we take \( t = -4 \).

\[ \alpha = -7,\quad \beta = 11,\quad \gamma = -4 \]

Step 2: Identify direction vectors and points on the lines.

For the first line:
\[ \vec{d}_1 = \langle 1,2,3 \rangle,\quad P_1(-7,11,-4) \]
For the second line:
\[ \vec{d}_2 = \langle 2,1,1 \rangle,\quad P_2(-5,10,3) \]

Step 3: Use the shortest distance formula between two skew lines.

\[ \text{Shortest distance} = \frac{|(\vec{P_2P_1}) \cdot (\vec{d}_1 \times \vec{d}_2)|} {|\vec{d}_1 \times \vec{d}_2|} \]
\[ \vec{P_2P_1} = \langle -2,1,-7 \rangle \]
\[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \langle -1,5,-3 \rangle \]
\[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{35} \]
\[ |(\vec{P_2P_1}) \cdot (\vec{d}_1 \times \vec{d}_2)| = 28 \]

Step 4: Calculate the distance.

\[ \text{Shortest distance} = \frac{28}{\sqrt{35}} = 4\sqrt{\frac{7}{5}} \]

Final Answer:
\[ \boxed{4\sqrt{\frac{7}{5}}} \]
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