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Mathematics Question on Three Dimensional Geometry

Equation of line passing through the point (2,3,1)(2, 3, 1) and parallel to the line of intersection of the plane x−2y−z+5=0x- 2y - z + 5 = 0 and x+y+3z=6x + y + 3z = 6 is

A

x−25=y−3−4=z−13 \frac {x-2}{5}=\frac {y-3}{-4}=\frac {z-1}{3}

B

x−2−5=y−3−4=z−13 \frac {x-2}{-5}=\frac {y-3}{-4}=\frac {z-1}{3}

C

x−25=y−34=z−12 \frac {x-2}{5}=\frac {y-3}{4}=\frac {z-1}{2}

D

x−24=y−34=z−12 \frac {x-2}{4}=\frac {y-3}{4}=\frac {z-1}{2}

Answer

x−2−5=y−3−4=z−13 \frac {x-2}{-5}=\frac {y-3}{-4}=\frac {z-1}{3}

Explanation

Solution

Let the DR?s of line passing through intersection of the given two planes x - 2y - z + 5 = 0 and x + y + 3z = 6 are a, b and c.
Then, a−2b−c=0 a - 2b - c = 0
and a+b+3c=0a + b + 3c = 0
⇒a−6+1=−b3+1=c1+2\Rightarrow \frac{a}{-6+1}=\frac{-b}{3+1}=\frac{c}{1+2}
⇒a:b:c=−5:−4:3\Rightarrow a : b : c = -5 :-4 : 3
Since, required line passes through (2, 3, 1) and parallel to above line. Hence, its equation will be
x−2−5=y−3−4=z−13\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}