Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through P and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

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Aicharya · Accepted Answer

The d.r.s. of the normal to the plane are $1,-1,1$ $\therefore \quad$ The equation of line QM is $\begin{aligned} & \frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\lambda(\text { say }) \\ & \Rightarrow x=\lambda+3, y=-\lambda+1, z=\lambda+7 \end{aligned}$ Let $M \equiv(\lambda+3,-\lambda+1, \lambda+7)$ $\therefore \quad$ Equation of plane becomes $\begin{aligned} & 1(\lambda+3)-1(-\lambda+1)+1(\lambda+7)=3 \\ & \Rightarrow 3 \lambda+6=0 \Rightarrow \lambda=-2 \\ & \mathrm{M} \equiv(1,3,5) \end{aligned}$
Since $M$ is the midpoint of PQ . $\therefore \quad \frac{3+\mathrm{a}}{2}=1, \frac{1+\mathrm{b}}{2}=3, \frac{7+\mathrm{c}}{2}=5$ $\Rightarrow a=-1, b=5, c=3$
Equation of the plane passing through P and containing the given line is $\begin{aligned} & \left|\begin{array}{ccc} x+1 & y-5 & z-3 \\ 1 & -5 & -3 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Rightarrow x-4 y+7 z=0 \end{aligned}$

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the…

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