The equation of image of the line {x-1}/9 = \frac{y-2}{-1} = \frac{z+3}{-3} in the plane 3x - 3y + 10z = 26 0 is

Correct Answer is : x−5∕29 = y−1∕2−1 = z−2−3

{x-2}/9 = \frac{y-2}{-1} = \frac{z+3}{-3}

{x-5\/2}/9 = \frac{y-1\/2}{-1} = \frac{z-2}{-3}

{x-1}/9 = \frac{y-2}{-1} = \frac{z+3}{-3}

{x-4}/9 = \frac{y+1}{-1} = \frac{z-7}{-3}

JEE Mains Model Paper 13

Question : 42 of 90

Marks: +1, -0

The equation of image of the line

x−1

y−2

−1

z+3

−3

in the plane 3x - 3y + 10z = 26 0 is

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