1. Key Concepts and Formulas

This problem involves two main concepts from 3D Geometry:

1. Finding the Equation of a Plane: A plane can be uniquely determined by three non-collinear points. A common method is to find two vectors lying in the plane, calculate their cross product to get a normal vector to the plane, and then use one of the given points to find the constant term in the plane equation.

   • If $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is the normal vector to the plane, and $(x_1, y_1, z_1)$ is a point on the plane, the equation of the plane is given by: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$
   • This can be expanded to the general form: $Ax + By + Cz + D = 0$.

2. Distance of a Point from a Plane: The perpendicular distance $d$ of a point $P(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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2. Step-by-Step Solution

Step 1: Identify the Given Points We are given three points that lie on the plane:

• $A = (2, -3, 1)$
• $B = (-1, 1, -2)$
• $C = (3, -4, 2)$

And the point from which we need to find the distance to the plane:

• $P = (7, -3, -4)$

Step 2: Form Two Vectors Lying in the Plane To find the normal vector to the plane, we first need to define two non-parallel vectors that lie within the plane. We can do this by forming vectors from two pairs of the given points. Let's choose $\vec{AB}$ and $\vec{AC}$.

• Vector $\vec{AB}$: This vector goes from point $A$ to point $B$. $\vec{AB} = B - A = (-1 - 2)\hat{i} + (1 - (-3))\hat{j} + (-2 - 1)\hat{k}$ $\vec{AB} = -3\hat{i} + 4\hat{j} - 3\hat{k}$

• Vector $\vec{AC}$: This vector goes from point $A$ to point $C$. $\vec{AC} = C - A = (3 - 2)\hat{i} + (-4 - (-3))\hat{j} + (2 - 1)\hat{k}$ $\vec{AC} = 1\hat{i} - 1\hat{j} + 1\hat{k} = \hat{i} - \hat{j} + \hat{k}$ Reasoning: The cross product of two non-parallel vectors lying in a plane yields a vector that is perpendicular (normal) to that plane. This normal vector is essential for defining the plane's equation.

Step 3: Determine the Normal Vector to the Plane The normal vector $\vec{n}$ to the plane is obtained by taking the cross product of $\vec{AB}$ and $\vec{AC}$. $\vec{n} = \vec{AB} \times \vec{AC}$ We calculate this using the determinant form: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix}$ Expanding the determinant: $\vec{n} = \hat{i}((4)(1) - (-3)(-1)) - \hat{j}((-3)(1) - (-3)(1)) + \hat{k}((-3)(-1) - (4)(1))$ $\vec{n} = \hat{i}(4 - 3) - \hat{j}(-3 - (-3)) + \hat{k}(3 - 4)$ $\vec{n} = \hat{i}(1) - \hat{j}(0) + \hat{k}(-1)$ $\vec{n} = \hat{i} - \hat{k}$ So, the normal vector to the plane is $\vec{n} = (1, 0, -1)$. This means the coefficients for $x, y, z$ in the plane equation are $A=1$, $B=0$, and $C=-1$. Reasoning: The components of the normal vector $(A, B, C)$ directly give the coefficients of $x, y, z$ in the general equation of the plane $Ax + By + Cz + D = 0$.

Step 4: Find the Equation of the Plane We use the normal vector $\vec{n} = (1, 0, -1)$ and one of the points lying on the plane, for example, $A(x_1, y_1, z_1) = (2, -3, 1)$. The equation of the plane is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$. Substituting the values: $1(x - 2) + 0(y - (-3)) + (-1)(z - 1) = 0$ $x - 2 + 0 - z + 1 = 0$ $x - z - 1 = 0$ This is the equation of the plane. From this, we can identify $A=1, B=0, C=-1, D=-1$. Reasoning: Having the normal vector and a point on the plane is sufficient to write down the complete algebraic equation of the plane.

Step 5: Calculate the Distance of Point P from the Plane Now we need to find the distance of the point $P(x_0, y_0, z_0) = (7, -3, -4)$ from the plane $x - z - 1 = 0$. Using the distance formula: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Substitute $A=1, B=0, C=-1, D=-1$ and $(x_0, y_0, z_0) = (7, -3, -4)$: $d = \frac{|(1)(7) + (0)(-3) + (-1)(-4) + (-1)|}{\sqrt{1^2 + 0^2 + (-1)^2}}$ $d = \frac{|7 + 0 + 4 - 1|}{\sqrt{1 + 0 + 1}}$ $d = \frac{|10|}{\sqrt{2}}$ $d = \frac{10}{\sqrt{2}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$: $d = \frac{10\sqrt{2}}{2}$ $d = 5\sqrt{2}$ Reasoning: This formula directly calculates the perpendicular distance, which is the shortest distance, from a given point to a plane.

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3. Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with signs when calculating the determinant for the cross product. A single sign error can lead to an incorrect normal vector and thus an incorrect plane equation.
• Coordinate Substitution: Ensure correct substitution of coordinates ($x_1, y_1, z_1$ for a point on the plane and $x_0, y_0, z_0$ for the point whose distance is being calculated) into the respective formulas.
• Absolute Value: Remember the absolute value in the distance formula, as distance is always non-negative.
• Rationalizing Denominators: It's good practice to rationalize the denominator of the final answer if it contains a square root.

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4. Summary

To find the distance of a point from a plane defined by three points, we first determined two vectors lying in the plane from the given three points. Their cross product yielded the normal vector to the plane. Using this normal vector and one of the points, we formulated the equation of the plane. Finally, we applied the formula for the distance of a point from a plane to find the required distance. The calculated distance is $5\sqrt{2}$.

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5. Final Answer The final answer is $\boxed{5\sqrt{2}}$, which corresponds to option (D).