1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $\vec{a}$ and parallel to a direction vector $\vec{v}$ can be represented as $\vec{r} = \vec{a} + t\vec{v}$, where $t$ is a scalar parameter.
• Line of Intersection of Two Planes: The line of intersection of two planes $P_1: \vec{n_1} \cdot \vec{r} = d_1$ and $P_2: \vec{n_2} \cdot \vec{r} = d_2$ has a direction vector $\vec{v}$ that is perpendicular to both normal vectors $\vec{n_1}$ and $\vec{n_2}$. Thus, $\vec{v}$ can be found by taking the cross product $\vec{n_1} \times \vec{n_2}$. To find a point on the line, we can set one coordinate to zero (e.g., $z=0$) and solve the resulting system of two linear equations for the other two coordinates.
• Distance from a Point to a Line: The shortest distance $d$ from a point $P$ to a line passing through a point $A$ with direction vector $\vec{v}$ is given by the formula: $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$

2. Step-by-Step Solution

Let the given point be $P(-1, 2, -2)$. The line $L$ is the intersection of two planes: Plane 1 ($P_1$): $2x + 3y + 2z = 0$ Plane 2 ($P_2$): $x - 2y + z = 0$

Step 1: Determine the Equation of the Line of Intersection ($L$)

To find the equation of the line, we need a point on the line and its direction vector.

• 1.1 Find the Direction Vector ($\vec{v}$) of the Line: The direction vector of the line of intersection is perpendicular to the normal vectors of both planes. The normal vector to $P_1$ is $\vec{n_1} = 2\hat{i} + 3\hat{j} + 2\hat{k}$. The normal vector to $P_2$ is $\vec{n_2} = \hat{i} - 2\hat{j} + \hat{k}$. The direction vector $\vec{v}$ is given by the cross product of $\vec{n_1}$ and $\vec{n_2}$: $\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ 1 & -2 & 1 \end{vmatrix}$ $\vec{v} = \hat{i}((3)(1) - (2)(-2)) - \hat{j}((2)(1) - (2)(1)) + \hat{k}((2)(-2) - (3)(1))$ $\vec{v} = \hat{i}(3 + 4) - \hat{j}(2 - 2) + \hat{k}(-4 - 3)$ $\vec{v} = 7\hat{i} - 0\hat{j} - 7\hat{k} = (7, 0, -7)$ We can use a simpler parallel vector by dividing by 7: $\vec{v} = (1, 0, -1)$.

• 1.2 Find a Point ($A$) on the Line: To find a point on the line of intersection, we can set one variable to zero and solve the system of equations. Let's set $z=0$: From $P_1$: $2x + 3y + 2(0) = 0 \Rightarrow 2x + 3y = 0$ (Equation 1) From $P_2$: $x - 2y + (0) = 0 \Rightarrow x - 2y = 0$ (Equation 2) From Equation 2, $x = 2y$. Substitute this into Equation 1: $2(2y) + 3y = 0$ $4y + 3y = 0$ $7y = 0 \Rightarrow y = 0$ Since $y=0$ and $x=2y$, we have $x=0$. Thus, a point on the line is $A(0, 0, 0)$.

  The equation of the line $L$ can be written as $\vec{r} = (0, 0, 0) + t(1, 0, -1)$, or $\vec{r} = (t, 0, -t)$.

Step 2: Calculate the Distance from Point $P$ to the Line $L$

We use the formula $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$. The given point is $P(-1, 2, -2)$. The point on the line is $A(0, 0, 0)$. The direction vector of the line is $\vec{v} = (1, 0, -1)$.

• 2.1 Calculate the Vector $\vec{AP}$: $\vec{AP} = P - A = (-1 - 0, 2 - 0, -2 - 0) = (-1, 2, -2)$

• 2.2 Calculate the Cross Product $\vec{AP} \times \vec{v}$: $\vec{AP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -2 \\ 1 & 0 & -1 \end{vmatrix}$ $= \hat{i}((2)(-1) - (-2)(0)) - \hat{j}((-1)(-1) - (-2)(1)) + \hat{k}((-1)(0) - (2)(1))$ $= \hat{i}(-2 - 0) - \hat{j}(1 + 2) + \hat{k}(0 - 2)$ $= -2\hat{i} - 3\hat{j} - 2\hat{k} = (-2, -3, -2)$

• 2.3 Calculate the Magnitudes: $|\vec{AP} \times \vec{v}| = \sqrt{(-2)^2 + (-3)^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$ $|\vec{v}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$

• 2.4 Calculate the Distance $d$: $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|} = \frac{\sqrt{17}}{\sqrt{2}}$ To rationalize the denominator: $d = \frac{\sqrt{17} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{34}}{2}$

3. Common Mistakes & Tips

• Direction Vector Calculation: A common error is incorrectly calculating the cross product of the normal vectors. Double-check the signs and terms.
• Point on the Line: Ensure the chosen point actually satisfies both plane equations. Setting one variable to zero is a straightforward way to find such a point, especially when the planes pass through the origin.
• Distance Formula: Using the correct distance formula is crucial. Be careful with vector operations (dot product vs. cross product) and magnitudes.
• Alternative Method (Foot of Perpendicular): While the cross product method is often quicker, finding the foot of the perpendicular $Q$ from $P$ to $L$ and then calculating $PQ$ is an equally valid approach. If time permits, using both methods can serve as a cross-check. For this problem, it would involve taking a general point $Q(t, 0, -t)$ on the line, forming $\vec{PQ}$, and setting $\vec{PQ} \cdot \vec{v} = 0$ to find $t$. This would lead to $t = 1/2$, giving $Q(1/2, 0, -1/2)$, and then $PQ = \sqrt{(3/2)^2 + (-2)^2 + (3/2)^2} = \sqrt{9/4 + 16/4 + 9/4} = \sqrt{34/4} = \frac{\sqrt{34}}{2}$. This confirms the result.

4. Summary

First, we determined the equation of the line of intersection by finding its direction vector (cross product of normal vectors of the planes) and a point on the line (by setting $z=0$ in the plane equations). The line passes through the origin $(0,0,0)$ and has a direction vector $(1,0,-1)$. Then, we calculated the distance from the given point $P(-1,2,-2)$ to this line using the formula $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$. The calculations yielded a distance of $\frac{\sqrt{34}}{2}$.

5. Final Answer

The final answer is $\boxed{\frac{\sqrt{34}}{2}}$, which corresponds to option (D).