Key Concepts and Formulas

1. Equation of a Plane: The equation of a plane passing through a point $(x_1, y_1, z_1)$ with a normal vector $\vec{n} = (a, b, c)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Here, $(a, b, c)$ are the direction ratios of the normal vector.
2. Perpendicular Planes: Two planes are perpendicular if and only if their normal vectors are perpendicular. If $\vec{n_1} = (A_1, B_1, C_1)$ and $\vec{n_2} = (A_2, B_2, C_2)$ are the normal vectors to two planes, then for them to be perpendicular, their dot product must be zero: $\vec{n_1} \cdot \vec{n_2} = A_1A_2 + B_1B_2 + C_1C_2 = 0$.
3. Distance of a Point from a Plane: The perpendicular distance $d$ of a point $P_0(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Step-by-Step Solution

Our primary goal is to first determine the equation of the required plane and then use the distance formula to find the distance of the given point from it.

Step 1: Formulate the General Equation of the Required Plane

• What we're doing: We are setting up a general algebraic expression for our target plane using the given point it passes through and an unknown normal vector.
• Why we're doing it: The problem states the plane passes through the point $(1, 2, 2)$. If we let the normal vector to this plane be $\vec{n} = (a, b, c)$, then its equation can be written directly using the point-normal form.
• Let the required plane be $\Pi$. Its equation is: $a(x - 1) + b(y - 2) + c(z - 2) = 0 \quad \text{(Equation 1)}$

Step 2: Apply the Perpendicularity Condition with the First Given Plane

• What we're doing: We are using the condition that the required plane is perpendicular to the first given plane to establish a relationship between the components of our unknown normal vector $(a, b, c)$.
• Why we're doing it: The normal vector of a plane is perpendicular to the plane itself. If two planes are perpendicular, their respective normal vectors must also be perpendicular. This means their dot product is zero.
• The normal vector to our required plane $\Pi$ is $\vec{n} = (a, b, c)$.
• The first given plane is $\Pi_1: x - y + 2z = 3$. Its normal vector is $\vec{n_1} = (1, -1, 2)$.
• Since $\Pi \perp \Pi_1$, we have $\vec{n} \cdot \vec{n_1} = 0$: $a(1) + b(-1) + c(2) = 0$ $a - b + 2c = 0 \quad \text{(Equation 2)}$

Step 3: Apply the Perpendicularity Condition with the Second Given Plane

• What we're doing: Similar to Step 2, we use the second perpendicularity condition to obtain another independent relationship for $(a, b, c)$.
• Why we're doing it: Having two such equations allows us to solve for the ratios of $a, b, c$, which are sufficient to define the direction of the normal vector.
• The second given plane is $\Pi_2: 2x - 2y + z + 12 = 0$. Its normal vector is $\vec{n_2} = (2, -2, 1)$.
• Since $\Pi \perp \Pi_2$, we have $\vec{n} \cdot \vec{n_2} = 0$: $a(2) + b(-2) + c(1) = 0$ $2a - 2b + c = 0 \quad \text{(Equation 3)}$

Step 4: Determine the Direction Ratios of the Normal Vector $(a, b, c)$

• What we're doing: We are solving the system of linear equations from Step 2 and Step 3 to find the values of $a, b, c$ (or their ratios).
• Why we're doing it: Once we have the direction ratios of the normal vector, we can substitute them back into Equation 1 to get the specific equation of the required plane.
• We have the system:
  1. $a - b + 2c = 0$
  2. $2a - 2b + c = 0$
• We can use the cross-multiplication method to solve for $a, b, c$: $\frac{a}{(-1)(1) - (2)(-2)} = \frac{b}{(2)(2) - (1)(1)} = \frac{c}{(1)(-2) - (-1)(2)}$ $\frac{a}{-1 + 4} = \frac{b}{4 - 1} = \frac{c}{-2 + 2}$ $\frac{a}{3} = \frac{b}{3} = \frac{c}{0}$
• Let this common ratio be $k$ (where $k \ne 0$, as $(0,0,0)$ cannot be a normal vector). Then, we can set $a = 3k$, $b = 3k$, and $c = 0k = 0$.
• For simplicity, we choose $k=1$. Thus, we can take the normal vector $\vec{n} = (3, 3, 0)$.

Step 5: Formulate the Equation of the Required Plane

• What we're doing: We are substituting the determined direction ratios of the normal vector into the general plane equation from Step 1.
• Why we're doing it: This gives us the complete, explicit equation of the plane whose distance from the given point we need to find.
• Substitute $a=3, b=3, c=0$ into Equation 1: $3(x - 1) + 3(y - 2) + 0(z - 2) = 0$ $3x - 3 + 3y - 6 = 0$ $3x + 3y - 9 = 0$
• Dividing the entire equation by 3 to simplify: $x + y - 3 = 0 \quad \text{(Equation of the Required Plane)}$

Step 6: Calculate the Distance of the Point from the Plane

• What we're doing: We are applying the point-to-plane distance formula using the coordinates of the given point and the equation of the plane we just found.
• Why we're doing it: This is the final step that directly answers the question asked in the problem.
• We need to find the distance of the point $P_0(1, -2, 4)$ from the plane $x + y - 3 = 0$.
• Using the distance formula $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$: Here, $(x_0, y_0, z_0) = (1, -2, 4)$ and the plane is $1x + 1y + 0z - 3 = 0$, so $A=1, B=1, C=0, D=-3$. $d = \frac{|1(1) + 1(-2) + 0(4) - 3|}{\sqrt{1^2 + 1^2 + 0^2}}$ $d = \frac{|1 - 2 + 0 - 3|}{\sqrt{1 + 1 + 0}}$ $d = \frac{|-4|}{\sqrt{2}}$ $d = \frac{4}{\sqrt{2}}$
• Rationalizing the denominator: $d = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

Common Mistakes & Tips

• Interpreting $c/0$ in Cross-Multiplication: When using the cross-multiplication method and you get a denominator of zero for one variable (e.g., $c/0$), it implies that the numerator for that variable must also be zero for the equality to hold (e.g., $c=0$). This means that component of the vector is zero.
• Choosing the Normal Vector: Remember that $(a, b, c)$ represents a direction. Any non-zero scalar multiple of $(a, b, c)$ will define the same normal direction and thus the same plane. Choosing a simple multiple (like $k=1$) simplifies calculations without affecting the final plane equation or distance.
• Simplifying Plane Equation: After finding the plane equation, always simplify it by dividing by common factors if possible. This makes subsequent calculations (like the distance formula) easier.

Summary

To find the distance of a point from a plane, we first needed to determine the equation of that plane. We achieved this by using the given point the plane passes through to form a general equation with an unknown normal vector $(a,b,c)$. The two perpendicularity conditions provided a system of linear equations, which we solved using cross-multiplication to find the direction ratios of the normal vector. Substituting these ratios back into the general equation yielded the specific equation of the plane. Finally, we applied the standard formula for the distance of a point from a plane to arrive at the solution.

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