Question

The distance of the point (1, 1, 9) from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x + y + z = 17$ is :

• A. $2\sqrt{19}$
• B. $19\sqrt{2}$
• C. $\sqrt{38}$
• D. 38

Hint:

Always use the parametric form of a line when looking for its intersection with a plane.

Answer 1

The Correct Option is C

Step 1: General point on the line: $(r+3, 2r+4, 2r+5)$.
Step 2: Substitute in plane: $(r+3) + (2r+4) + (2r+5) = 17 \Rightarrow 5r = 5 \Rightarrow r = 1$.
Step 3: Intersection point: $(4, 6, 7)$.
Step 4: Distance from (1, 1, 9): $d = \sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.