Question

Let \( L \) be the line \[ \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} \] and let \( S \) be the set of all points \( (a,b,c) \) on \( L \), whose distance from the line \[ \frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0} \] along the line \( L \) is \( 7 \). Then \[ \sum_{(a,b,c)\in S} (a+b+c) \] is equal to

• A. 34
• B. 40
• C. 6
• D. 28

Hint:

For problems involving distance along a line, always use the unit direction vector of the line and move forward and backward by the given distance.

Answer 1

The Correct Option is B

Step 1: Write the parametric form of line \( L \).
From \[ \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} = t, \] we get \[ x = 2t - 1,\quad y = 3t - 1,\quad z = 6t - 3. \]
Step 2: Find the point of intersection with the given second line.
The second line is \[ \frac{x+1}{2} = \frac{y+1}{3}, \quad z = 9. \] Substitute \( z = 9 \) in the parametric form of \( L \): \[ 6t - 3 = 9 \Rightarrow t = 2. \] Thus, the point of intersection is \[ P(3, 5, 9). \]
Step 3: Find points at distance 7 along the line \( L \).
Direction ratios of \( L \) are \( (2,3,6) \). Magnitude: \[ \sqrt{2^2 + 3^2 + 6^2} = 7. \] Hence, unit direction vector is \[ \left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right). \] Points at distance \( 7 \) from \( P \) along \( L \) are \[ P \pm 7\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right). \] Thus, the two points are: \[ P_1 = (3,5,9) + (2,3,6) = (5,8,15), \] \[ P_2 = (3,5,9) - (2,3,6) = (1,2,3). \]
Step 4: Compute the required sum.
\[ (a+b+c)_{P_1} = 5 + 8 + 15 = 28, \] \[ (a+b+c)_{P_2} = 1 + 2 + 3 = 6. \] \[ \sum_{(a,b,c)\in S} (a+b+c) = 28 + 6 = 40. \]
Final Answer: \[ \boxed{40} \]