Key Concepts and Formulas

• Equation of a Plane: A plane can be defined by a point $(x_0, y_0, z_0)$ lying on it and a vector $\vec{n} = \langle A, B, C \rangle$ perpendicular to it (called the normal vector). The equation of the plane is given by $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, which can be expanded to the general form $Ax+By+Cz+D=0$.
• Normal Vector from Two Vectors in the Plane: If two non-parallel vectors $\vec{u}$ and $\vec{v}$ lie in a plane, their cross product $\vec{u} \times \vec{v}$ yields a vector perpendicular to both, which serves as a normal vector $\vec{n}$ to the plane.
• Distance of a Plane from the Origin: The perpendicular distance $d$ of a plane $Ax+By+Cz+D=0$ from the origin $(0,0,0)$ is given by the formula: $d = \frac{|A(0) + B(0) + C(0) + D|}{\sqrt{A^2+B^2+C^2}} = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$

Step-by-Step Solution

Step 1: Identify Points and Vectors Lying within the Plane

• What we are doing: To find the equation of a plane, we need a point on the plane and its normal vector. To find the normal vector, we typically need two non-parallel vectors that lie within the plane.
• Reasoning: The problem states that plane P passes through a specific point and contains a given line. These pieces of information allow us to extract the necessary points and vectors.
• We are given a point $A(3, 7, -7)$ that lies on the plane P.
• The given line is $L: \frac{x - 2}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.
  • From the line equation, we can identify a point on the line: $B(2, 3, -2)$. Since the line lies in plane P, point $B$ also lies in plane P.
  • The direction ratios of the line are $\langle -3, 2, 1 \rangle$. Thus, the direction vector of line $L$ is $\vec{v} = \langle -3, 2, 1 \rangle$. This vector lies entirely within plane P.
• Now we have two points on the plane, $A(3, 7, -7)$ and $B(2, 3, -2)$. We can form a vector connecting these two points, which also lies in the plane: $\vec{AB} = B - A = \langle 2-3, 3-7, -2-(-7) \rangle = \langle -1, -4, 5 \rangle$ We now have two vectors lying in the plane: $\vec{v} = \langle -3, 2, 1 \rangle$ and $\vec{AB} = \langle -1, -4, 5 \rangle$.

Step 2: Determine the Normal Vector to the Plane

• What we are doing: The normal vector $\vec{n}$ is perpendicular to the plane. We can find it by taking the cross product of any two non-parallel vectors that lie in the plane.
• Reasoning: The cross product of two vectors results in a vector that is orthogonal to both input vectors. Since $\vec{v}$ and $\vec{AB}$ are in the plane, their cross product will be perpendicular to the plane, thus serving as our normal vector.
• Let's calculate the cross product of $\vec{v}$ and $\vec{AB}$: $\vec{n} = \vec{v} \times \vec{AB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 2 & 1 \\ -1 & -4 & 5 \end{vmatrix}$ $\vec{n} = \mathbf{i}((2)(5) - (1)(-4)) - \mathbf{j}((-3)(5) - (1)(-1)) + \mathbf{k}((-3)(-4) - (2)(-1))$ $\vec{n} = \mathbf{i}(10 + 4) - \mathbf{j}(-15 + 1) + \mathbf{k}(12 + 2)$ $\vec{n} = 14\mathbf{i} - (-14)\mathbf{j} + 14\mathbf{k}$ $\vec{n} = 14\mathbf{i} + 14\mathbf{j} + 14\mathbf{k}$
• The direction ratios of the normal vector are $\langle 14, 14, 14 \rangle$. For simplicity, we can use any non-zero scalar multiple of this vector. Dividing by 14, we choose $\vec{n}' = \langle 1, 1, 1 \rangle$ as our simplified normal vector.

Step 3: Formulate the Equation of the Plane P

• What we are doing: We use the point-normal form of the plane equation to get the algebraic representation of plane P.
• Reasoning: We have a point on the plane (e.g., $A(3, 7, -7)$) and the normal vector $\vec{n}' = \langle 1, 1, 1 \rangle$. These are the two essential components for the plane equation.
• Substitute the point $A(x_0, y_0, z_0) = (3, 7, -7)$ and the normal vector $\vec{n}' = \langle A, B, C \rangle = \langle 1, 1, 1 \rangle$ into the point-normal form: $1(x - 3) + 1(y - 7) + 1(z - (-7)) = 0$ $x - 3 + y - 7 + z + 7 = 0$ $x + y + z - 3 = 0$
• So, the equation of plane P is $x + y + z - 3 = 0$.

Step 4: Calculate the Distance of the Plane P from the Origin

• What we are doing: We apply the formula for the distance of a plane from the origin.
• Reasoning: The problem explicitly asks for this distance, $d$.
• From our plane equation $x + y + z - 3 = 0$, we identify the coefficients $A=1, B=1, C=1,$ and $D=-3$.
• Substitute these values into the distance formula: $d = \frac{|D|}{\sqrt{A^2+B^2+C^2}} = \frac{|-3|}{\sqrt{1^2 + 1^2 + 1^2}}$ $d = \frac{3}{\sqrt{1 + 1 + 1}}$ $d = \frac{3}{\sqrt{3}}$ $d = \sqrt{3}$

Step 5: Calculate $d^2$

• What we are doing: The question specifically asks for the value of $d^2$.
• Reasoning: This is the final calculation required by the problem.
• We found $d = \sqrt{3}$.
• Therefore, $d^2 = (\sqrt{3})^2 = 3$.

Common Mistakes & Tips

• Vector Order in Cross Product: While $\vec{u} \times \vec{v}$ gives a normal vector, $\vec{v} \times \vec{u}$ gives a normal vector in the opposite direction. Both are valid normal vectors for the same plane, as they only change the sign of the plane equation ($Ax+By+Cz+D=0$ vs. $-Ax-By-Cz-D=0$), representing the identical plane.
• Simplifying Normal Vector: Always simplify the direction ratios of the normal vector by dividing by their greatest common divisor. This makes subsequent calculations (like the plane equation and distance formula) much easier.
• Careful with Signs: Pay close attention to negative signs, especially when calculating determinants for the cross product and when plugging values into the plane equation and distance formula.
• Alternative Method (Determinant Form): The equation of a plane containing a line $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ and passing through a point $(x_2, y_2, z_2)$ can be found directly using a determinant: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ l & m & n \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \end{vmatrix} = 0$ This method combines Steps 1, 2, and 3 into a single calculation. Using this method for the given problem also yields $x+y+z-3=0$.

Summary

To find the distance of plane P from the origin, we first determined two vectors lying within the plane: the direction vector of the given line and a vector connecting the given point to a point on the line. Their cross product gave us the normal vector $\langle 1, 1, 1 \rangle$. Using this normal vector and one of the points on the plane, we formulated the equation of the plane as $x+y+z-3=0$. Finally, we applied the distance formula from the origin to this plane, which resulted in $d = \sqrt{3}$. Squaring this value, we found $d^2 = 3$.

The final answer is $\boxed{3}$.