Question

If \(P(1,2,5)\), \(Q(3,0,7)\), \(R(6,-3,10)\) are collinear and \((\alpha,\beta,\gamma)\) is a point at distance 3 from \(P\) on the same line, then the value of \(\alpha+\beta+\gamma\) lying between 6 and 7 is:

• A. \(13-\sqrt3\)
• B. \(8-\sqrt3\)
• C. \(16+\sqrt3\)
• D. \(7-\sqrt2\)

Hint:

To locate a point at a given distance on a line, use the unit direction vector.

Answer 1

The Correct Option is B

Concept: Use the unit vector along the line and move a distance 3 from point \(P\).

Step 1: Find direction vector.
\[ Q-P=(2,-2,2). \] Unit vector \[ \frac{(1,-1,1)}{\sqrt3}. \]

Step 2: Move 3 units from \(P\).
Required point \[ P\pm3\left(\frac{1,-1,1}{\sqrt3}\right). \] \[ = (1,2,5)\pm(\sqrt3,-\sqrt3,\sqrt3). \]

Step 3: Find sum of coordinates.
\[ \alpha+\beta+\gamma = 8\pm\sqrt3. \] Value between 6 and 7 is \[ \boxed{8-\sqrt3}. \]