Let P be the image of the point (3, 1, 7) wit | vedclass

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(C) Let the given point be $Q(3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane $x-y+z=3$ has the direction ratios $(1, -1, 1)$.I

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$x-4y+7z=0$

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(C) Let the given point be $Q(3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane $x-y+z=3$ has the direction ratios $(1, -1, 1)$.Its equation is $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.Any point on this line is $(3+\lambda, 1-\lambda, 7+\lambda)$.This point lies on the plane $x-y+z=3$,so $(3+\lambda) - (1-\lambda) + (7+\lambda) = 3$.$3+\lambda-1+\lambda+7+\lambda=3 \Rightarrow 3\lambda+9=3 \Rightarrow 3\lambda=-6 \Rightarrow \lambda=-2$.The foot of the perpendicular $R$ is $(3-2, 1-(-2), 7-2) = (1, 3, 5)$.Let $P(x_1, y_1, z_1)$ be the image of $Q$. Since $R$ is the midpoint of $PQ$,we have $\frac{x_1+3}{2}=1, \frac{y_1+1}{2}=3, \frac{z_1+7}{2}=5$.$x_1 = -1, y_1 = 5, z_1 = 3$. So $P$ is $(-1, 5, 3)$.The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$.The plane equation is $a(x+1) + b(y-5) + c(z-3) = 0$.Since it contains the line,it passes through $(0, 0, 0)$,so $a(1) + b(-5) + c(-3) = 0 \Rightarrow a-5b-3c=0$.Also,the normal vector $(a, b, c)$ is perpendicular to the line direction $(1, 2, 1)$,so $a+2b+c=0$.Solving $a-5b-3c=0$ and $a+2b+c=0$ by subtraction: $-7b-4c=0 \Rightarrow b = -4c/7$.Then $a = 5(-4c/7) + 3c = -20c/7 + 21c/7 = c/7$.Taking $c=7$,we get $a=1, b=-4$. The equation is $1(x+1) - 4(y-5) + 7(z-3) = 0$.$x+1-4y+20+7z-21 = 0 \Rightarrow x-4y+7z=0$.

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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