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

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$\frac{x - 4}{5} = \frac{y + 1}{- 3} = \frac{z - 7}{- 1}$

$\frac{x - 4}{7} = \frac{y + 1}{- 5} = \frac{z - 7}{- 3}$

$\frac{x - 4}{- 3} = \frac{y + 1}{- 5} = \frac{z - 7}{- 1}$

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

answer is D.

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Detailed Solution

Given line is

$\frac{x - 1}{9} = \frac{y - 2}{- 1} = \frac{z + 3}{- 3}$ (i)

and the plane is

$3 x - 3 y + 10 z = 26$

The direction ratios of the line are 9, –1, – 3 and the direction ratios of the normal to the given plane are 3, –3 and 10. Clearly line (i) is parallel to the plane (ii).

Let Q be the image of the point P(1, 2, –3). Consider $Q = (α, β, γ)$

Now,

$\begin{array}{l} \frac{α - 1}{3} = \frac{β - 2}{- 3} = \frac{γ + 3}{10} = - 2 (\frac{3 - 6 - 30 - 26}{9 + 9 + 100}) \\ \frac{α - 1}{3} = \frac{β - 2}{- 3} = \frac{γ + 3}{10} = 1 \\ α = 4, β = - 1, γ = 7 \end{array}$