Key Concepts and Formulas

1. Direction Vector of a Line: For a line given in symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$, its direction vector is $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k} = \langle a, b, c \rangle$.
2. Normal Vector to a Plane (using Cross Product): If a plane's normal vector $\vec{n}$ is perpendicular to two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$, then $\vec{n}$ is parallel to the cross product of these direction vectors. We can take $\vec{n} = \vec{d_1} \times \vec{d_2}$.
3. Equation of a Plane (Point-Normal Form): The equation of a plane passing through a point $(x_0, y_0, z_0)$ and having a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. This can be expanded to the general form $Ax + By + Cz + D = 0$.
4. Distance of a Point from a Plane: The perpendicular distance $D$ of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D_{plane} = 0$ is given by the formula: $D = \frac{|Ax_1 + By_1 + Cz_1 + D_{plane}|}{\sqrt{A^2 + B^2 + C^2}}$

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Step-by-Step Solution

Step 1: Identify the Direction Vectors of the Given Lines

To find the equation of the plane, we first need its normal vector. The problem states that the normal vector is perpendicular to two given lines. We extract the direction vectors from the symmetric equations of these lines.

• For Line 1: $\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 4}{3}$ The direction vector is $\vec{d_1} = \langle 1, -2, 3 \rangle$.
• For Line 2: $\frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z + 7}{-1}$ The direction vector is $\vec{d_2} = \langle 2, -1, -1 \rangle$.

Step 2: Determine the Normal Vector of the Plane

The normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. The cross product of two vectors yields a vector that is orthogonal to both of them. Therefore, we can find the normal vector $\vec{n}$ by calculating the cross product of $\vec{d_1}$ and $\vec{d_2}$:

$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix}$

Now, we expand the determinant: $\vec{n} = \hat{i}((-2)(-1) - (3)(-1)) - \hat{j}((1)(-1) - (3)(2)) + \hat{k}((1)(-1) - (-2)(2))$ $\vec{n} = \hat{i}(2 - (-3)) - \hat{j}(-1 - 6) + \hat{k}(-1 - (-4))$ $\vec{n} = \hat{i}(2 + 3) - \hat{j}(-7) + \hat{k}(-1 + 4)$ $\vec{n} = 5\hat{i} + 7\hat{j} + 3\hat{k}$ So, the normal vector to the plane is $\vec{n} = \langle 5, 7, 3 \rangle$. This means the coefficients for the plane equation are $A=5$, $B=7$, and $C=3$.

Step 3: Formulate the Equation of the Plane

We have the normal vector $\vec{n} = \langle 5, 7, 3 \rangle$ and a point that the plane passes through, $P_0 = (1, -1, -1)$. We use the point-normal form of the plane equation: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.

Substitute the values $A=5, B=7, C=3$ and $(x_0, y_0, z_0) = (1, -1, -1)$: $5(x - 1) + 7(y - (-1)) + 3(z - (-1)) = 0$ $5(x - 1) + 7(y + 1) + 3(z + 1) = 0$

Now, expand and simplify to the general form $Ax + By + Cz + D_{plane} = 0$: $5x - 5 + 7y + 7 + 3z + 3 = 0$ $5x + 7y + 3z + (-5 + 7 + 3) = 0$ $5x + 7y + 3z + 5 = 0$ This is the equation of the required plane.

Step 4: Calculate the Distance of the Point (1, 3, -7) from the Plane

We need to find the perpendicular distance of the point $P_1 = (1, 3, -7)$ from the plane $5x + 7y + 3z + 5 = 0$. Using the distance formula $D = \frac{|Ax_1 + By_1 + Cz_1 + D_{plane}|}{\sqrt{A^2 + B^2 + C^2}}$:

From the plane equation, we have $A=5$, $B=7$, $C=3$, and $D_{plane}=5$. The coordinates of the point are $(x_1, y_1, z_1) = (1, 3, -7)$.

Substitute these values into the formula: $D = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}}$ Perform the calculations: $D = \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}}$ $D = \frac{|10|}{\sqrt{83}}$ $D = \frac{10}{\sqrt{83}}$

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Common Mistakes & Tips

• Cross Product Signs: Be meticulous with signs when calculating the cross product, especially for the $\hat{j}$ component, which conventionally has a negative sign in the determinant expansion.
• Arithmetic Errors: Double-check all arithmetic, particularly when simplifying the plane equation and evaluating the numerator and denominator of the distance formula.
• Absolute Value: Remember to take the absolute value of the numerator in the distance formula, as distance is always non-negative.

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Summary

The problem required us to find the distance of a point from a plane. We first determined the direction vectors of the two lines given. Then, we used their cross product to find the normal vector of the plane. With the normal vector and the point through which the plane passes, we formulated the equation of the plane. Finally, we applied the standard formula to calculate the perpendicular distance from the given point to this plane. The calculated distance is $\frac{10}{\sqrt{83}}$.

The final answer is $\boxed{\frac{10}{\sqrt{83}}}$, which corresponds to option (A).