Question

The line \(\ell\) passes through the point \((2,1,1)\) and is parallel to the plane \(x+y+2z = 18\). If line \(\ell\) intersects the line \(\frac{x+2}{3} = \frac{y+1}{-1} = \frac{z-2}{1}\), then equation of the line \(\ell\) is...

• A. \(\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-1}{-2}\)
• B. \(\frac{x-3}{1} = \frac{y-2}{3} = \frac{z-2}{-2}\)
• C. \(\frac{x-2}{3} = \frac{y-1}{-1} = \frac{z-1}{-1}\)
• D. \(\frac{x-3}{1} = \frac{y-4}{3} = \frac{z+1}{-2}\)

Hint:

Find the point on the given line using the parallel condition, then join it to the point.

Answer 1

The Correct Option is D

Step 1: Test the options:
Direction ratios must satisfy $l + m + 2n = 0$. A: $3 + 1 - 4 = 0$ passes, B: $1 + 3 - 4 = 0$ passes, C: $3 - 1 - 2 = 0$ passes, D: $1 + 3 - 4 = 0$ passes. So use the point and intersection.

Step 2: Intersection check:
Option D passes through $(2, 1, 1)$ when its parameter is $-1$... as $(3 - 1, 4 - 3, -1 + 2) = (2, 1, 1)$. At parameter $-2$ it reaches $(1, -2, 3)$, which lies on the given line at $s = 1$.
So option D passes through both $(2, 1, 1)$ and a point of the given line.
Options A and C give $(2,1,1) + \lambda(3,1,-2)$ and $(2,1,1) + \lambda(3,-1,-1)$, which do not meet the given line.

Final Answer:
The line is option (D). \[ \boxed{\frac{x-3}{1}=\frac{y-4}{3}=\frac{z+1}{-2}} \]