Question

The distance of the point \( (3,4,5) \) from the point of intersection of the line \[ \frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} \] and the plane \( x+y+z=2 \) is

• A. \( 6 \) units
• B. \( 13 \) units
• C. \( 10 \) units
• D. \( 7 \) units

Hint:

Always find the point of intersection first, then apply the distance formula.

Answer 1

The Correct Option is A

Step 1: Write parametric equations of the line.
\[ x = 3 + \lambda,\quad y = 4 + 2\lambda,\quad z = 5 + 2\lambda \]

Step 2: Substitute in the plane equation.
\[ (3+\lambda) + (4+2\lambda) + (5+2\lambda) = 2 \] \[ 12 + 5\lambda = 2 \Rightarrow \lambda = -2 \]

Step 3: Find the point of intersection.
\[ x=1,\; y=0,\; z=1 \]

Step 4: Find the distance from \( (3,4,5) \).
\[ \text{Distance} = \sqrt{(3-1)^2 + (4-0)^2 + (5-1)^2} \] \[ = \sqrt{4+16+16} = \sqrt{36} = 6 \]

Step 5: Conclusion.
\[ \boxed{6 \text{ units}} \]