Equation of Plane Passing Through Image Point and Given Line

To find the equation of the required plane, we first determine the coordinates of point $P$, which is the mirror image of $A(3, 1, 7)$ about the plane $x - y + z = 3$.

Step 1: Find the coordinates of point $P$

The formula for the image $(x_1, y_1, z_1)$ of a point $(x_0, y_0, z_0)$ with respect to the plane $ax + by + cz + d = 0$ is given by: $\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = \frac{z_1 - z_0}{c} = -2 \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}$

Here:

• $(x_0, y_0, z_0) = (3, 1, 7)$
• Plane equation: $x - y + z - 3 = 0 \implies a = 1, b = -1, c = 1, d = -3$

Substituting these values into the expression: $-2 \frac{1(3) - 1(1) + 1(7) - 3}{1^2 + (-1)^2 + 1^2} = -2 \frac{6}{3} = -4$

Now, we calculate the coordinates of $P(x_1, y_1, z_1)$: $\frac{x_1 - 3}{1} = -4 \implies x_1 = -1$ $\frac{y_1 - 1}{-1} = -4 \implies y_1 = 5$ $\frac{z_1 - 7}{1} = -4 \implies z_1 = 3$

Thus, point $P$ is $(-1, 5, 3)$.

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Step 2: Find the equation of the plane containing the line and point $P$

The given line is: $L: \frac{x}{1} = \frac{y}{2} = \frac{z}{1}$

Since the line $L$ passes through the origin $O(0, 0, 0)$, the required plane must also pass through $O(0, 0, 0)$.

The general equation of a plane passing through the origin is: $Ax + By + Cz = 0$

Since the plane contains line $L$, its normal vector $\vec{n} = (A, B, C)$ is perpendicular to the direction vector of the line $\vec{v} = (1, 2, 1)$: $A(1) + B(2) + C(1) = 0 \implies A + 2B + C = 0 \quad \text{--- (1)}$

Furthermore, the point $P(-1, 5, 3)$ lies on the plane, so its coordinates must satisfy the plane equation: $A(-1) + B(5) + C(3) = 0 \implies -A + 5B + 3C = 0 \quad \text{--- (2)}$

Adding equations $(1)$ and $(2)$: $(A + 2B + C) + (-A + 5B + 3C) = 0 \implies 7B + 4C = 0 \implies B = -\frac{4}{7}C$

Substitute $B$ into equation $(1)$: $A + 2\left(-\frac{4}{7}C\right) + C = 0 \implies A - \frac{8}{7}C + C = 0 \implies A = \frac{1}{7}C$

Thus, the ratio of the direction ratios of the normal vector is: $A : B : C = \frac{1}{7}C : -\frac{4}{7}C : C = 1 : -4 : 7$

Substituting $A = 1$, $B = -4$, and $C = 7$ into the plane equation $Ax + By + Cz = 0$: $x - 4y + 7z = 0$

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Conclusion

The equation of the plane is $x - 4y + 7z = 0$, which corresponds to Option (C).