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Mathematics Question on introduction to three dimensional geometry

The image of the line x−23=y+14=z−212\frac{x-2}{3}=\frac{y+1}{4} = \frac{z-2}{12} in the plane 2x−y+z+3=02x - y + z + 3 = 0 is the line

A

x+33=y−51=z−2−5\frac{x+3}{3}=\frac{y-5}{1} = \frac{z-2}{-5}

B

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

C

x−33=y+51=z−2−5\frac{x-3}{3}=\frac{y+5}{1} = \frac{z-2}{-5}

D

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

Answer

x+33=y−51=z−2−5\frac{x+3}{3}=\frac{y-5}{1} = \frac{z-2}{-5}

Explanation

Solution

Here, plane, line and its image are parallel to each other. So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line.
Here, plane and line are parallel to each other.
Equation of normal to the plane through the point (1, 3, 4) is
\hspace40mm \frac{x-1}{2}=\frac{y-3}{-1} = \frac{z-4}{1} = k \, \, \, \, \, [say]
Any point in this normal is (2k + 1, - k + 3,4 + k).
Then, (2k+1+12,3−k+32,4+k+42)\bigg(\frac{2k+1+1}{2},\frac{3-k+3}{2}, \frac{4 +k+4}{2}\bigg) lies on plane.
⇒2(k+1)−(6−k2)+(8+k2)+3=0  ⇒   k=−2\Rightarrow 2(k+1)-\bigg(\frac{6-k}{2}\bigg)+\bigg(\frac{8+k}{2}\bigg)+3 = 0 \, \, \Rightarrow \, \, \, k = -2
Hence, point through which this image pass is
        (2k+1,3−k,4+k)\, \, \, \, \, \, \, \, (2k + 1,3 - k, 4 + k)
i.e. [2(−2)+1,3+2,4−2]=(−3,5,2)i.e.\, [2(-2) + 1 ,3 + 2,4 - 2] = (-3 ,5 ,2 )
Hence, equation of image line is \hspace40mm\frac{x+3}{3}=\frac{y-5}{1}= \frac{z-2}{-5}