The image of the point (1,2,−1), on the plane containing the line x+1−3=y−32=z+21 and the point (0,7,−7) , is

Given x+1−3=y−32=z+21……..(1) and point (0,7,−7) The equation of the plane is x+1y−3z+214−5−321=0⇒(x+1)(4+10)−(y−3)(1−15)+(z+2)(2+12)=0⇒14(x+1)+14(y−3)+14(z+2)=0⇒x+y+z=0………...(2) The image of the point (1,2,−1) on the plane (2) is h−x1a=k−y1b=l−z1c=−2(ax1+by1+cz1+d)a2+b2+c2⇒h−1=−43⇒h=−13 And ⇒k−2=−43⇒k=23 And l+1=−43⇒l=−73∴ Image of (1,2,−1) is −13,23,−73

The image of the point (1,2,−1), on the plane containing the line x+1−3=y−32=z+21 and the point (0,7,−7) , is

Given x+1−3=y−32=z+21……..(1) and point (0,7,−7) The equation of the plane is x+1y−3z+214−5−321=0⇒(x+1)(4+10)−(y−3)(1−15)+(z+2)(2+12)=0⇒14(x+1)+14(y−3)+14(z+2)=0⇒x+y+z=0………...(2) The image of the point (1,2,−1) on the plane (2) is h−x1a=k−y1b=l−z1c=−2(ax1+by1+cz1+d)a2+b2+c2⇒h−1=−43⇒h=−13 And ⇒k−2=−43⇒k=23 And l+1=−43⇒l=−73∴ Image of (1,2,−1) is −13,23,−73

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$The image of the point (1, 2, - 1), on the plane containing the line \frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} and$
$the point (0, 7, - 7), is$

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$(\frac{- 1}{3}, \frac{- 7}{3}, \frac{1}{3})$

$(\frac{- 1}{3}, \frac{2}{3}, \frac{- 7}{3})$

$(\frac{- 1}{3}, 0, \frac{- 7}{3})$

$(\frac{- 1}{3}, \frac{2}{3}, \frac{7}{3})$

answer is B.

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Detailed Solution

$\begin{array}{l} Given \frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} \dots \dots . . (1) and point (0, 7, - 7) \\ The equation of the plane is |\begin{array}{ccc} x + 1 & y - 3 & z + 2 \\ 1 & 4 & - 5 \\ - 3 & 2 & 1 \end{array}| = 0 \\ \Rightarrow (x + 1) (4 + 10) - (y - 3) (1 - 15) + (z + 2) (2 + 12) = 0 \\ \Rightarrow 14 (x + 1) + 14 (y - 3) + 14 (z + 2) = 0 \\ \Rightarrow x + y + z = 0 \dots \dots \dots . . . (2) \end{array}$

$\begin{array}{l} The image of the point (1, 2, - 1) on the plane (2) is \\ \frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{l - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}} \\ \Rightarrow h - 1 = \frac{- 4}{3} \\ \Rightarrow h = \frac{- 1}{3} \end{array}$

$\begin{array}{l} And \Rightarrow k - 2 = - \frac{4}{3} \\ \Rightarrow k = \frac{2}{3} \\ And l + 1 = \frac{- 4}{3} \\ \Rightarrow l = - \frac{7}{3} \\ ∴ Image of (1, 2, - 1) is (- \frac{1}{3}, \frac{2}{3}, - \frac{7}{3}) \end{array}$