The equation of the plane passing through the image of the point $P (3, 5, 7)$ w.r.to the plane $2 x + y + z = 3$ and having the line $\frac{x - 1}{- 1} = \frac{y}{2} = \frac{z - 2}{- 3}$ is:

x - y - z + 1 = 0

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3 y + 2 z - 4 = 0

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19 x + 5 y - 3 z = 13

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x + 5 y + 3 z = 7

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Solution

The correct option is B $3 y + 2 z - 4 = 0$
Let $P (α, β, γ)$ be the image of the point $Q (3, 5, 7) = (x_{1}, y_{1}, z_{1})$ w.r.to the plane $a x + b y + c z + d = o = 2 x + y + z - 3$ ,
We know,
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - 2 \frac{a x_{1} + b y_{1} + c z_{1} + d}{a^{2} + b^{2} + c^{2}}$
$\frac{α - 3}{2} = \frac{β - 5}{1} = \frac{γ - 7}{1} = - 2 \frac{2 (3) + 1 (5) + 1 (7) - 3}{4 + 1 + 1} = - 2 \times \frac{15}{6} = - 5$
$α = - 7, β = 0, γ = 2$
Image is $P (- 7, 0, 2)$
Now required plane passes through $P (- 7, 0, 2)$ and contains the line $\frac{x - 1}{- 1} = \frac{y}{2} = \frac{z - 2}{- 3}$
So plane passes through $P (- 7, 0, 2)$ and $A (1, 0, 2)$
D'R's of $A P = (8, 0, 0)$
$∴$ Plane equation is
$∣ ∣ ∣ ∣ \begin{matrix} x - 1 & y & z - 2 8 & 0 & 0 - 1 & 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 3 y + 2 z - 4 = 0$

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Similar questions

(3, 1, 7)

x - y + z = 3

\frac{x}{1} = \frac{y}{2} = \frac{z}{1}

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

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

(2, 3, 1)

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

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

α x + β y + γ z = 24,

α + β + γ

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

2 x + y + z = 5

(4, - 1, 2)

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

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

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