1. Key Concepts and Formulas

• Equation of a Plane Passing Through Three Non-Collinear Points: If three non-collinear points $P_1(x_1, y_1, z_1)$, $P_2(x_2, y_2, z_2)$, and $P_3(x_3, y_3, z_3)$ lie on a plane, its equation can be found using the determinant form: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$
• Alternative Method using Normal Vector: The equation of a plane passing through a point $P_0(x_0, y_0, z_0)$ and having a normal vector $\vec{n}=(A, B, C)$ is given by $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$. The normal vector can be determined by taking the cross product of two non-parallel vectors lying in the plane (e.g., $\vec{P_0P_1} \times \vec{P_0P_2}$).
• Intercept Form of a Plane: The equation of a plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a$, $b$, and $c$ are the x-intercept, y-intercept, and z-intercept, respectively. To convert a general plane equation $Ax + By + Cz + D = 0$ to intercept form, rearrange it as $Ax + By + Cz = -D$, and then divide the entire equation by $-D$.

2. Step-by-Step Solution

Step 1: Identify the points on the plane The problem states that the plane passes through the point $P_1(-2, -2, 2)$ and contains the line joining points $P_2(1, -1, 2)$ and $P_3(1, 1, 1)$. This implies that all three points $P_1$, $P_2$, and $P_3$ lie on the plane. Let their coordinates be:

• $P_1(x_1, y_1, z_1) = (-2, -2, 2)$
• $P_2(x_2, y_2, z_2) = (1, -1, 2)$
• $P_3(x_3, y_3, z_3) = (1, 1, 1)$

Step 2: Determine two vectors lying in the plane To find the normal vector of the plane, we first need two non-parallel vectors lying within the plane. We can form these vectors by taking differences between the coordinates of the given points. Let's use $P_1$ as the reference point for these vectors: Vector $\vec{P_1P_2} = (x_2-x_1, y_2-y_1, z_2-z_1)$ $\vec{P_1P_2} = (1 - (-2), -1 - (-2), 2 - 2) = (3, 1, 0)$

Vector $\vec{P_1P_3} = (x_3-x_1, y_3-y_1, z_3-z_1)$ $\vec{P_1P_3} = (1 - (-2), 1 - (-2), 1 - 2) = (3, 3, -1)$

Step 3: Find the normal vector to the plane The normal vector $\vec{n}$ to the plane is perpendicular to any two vectors lying in the plane. We can find it by taking the cross product of $\vec{P_1P_2}$ and $\vec{P_1P_3}$. $\vec{n} = \vec{P_1P_2} \times \vec{P_1P_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 0 \\ 3 & 3 & -1 \end{vmatrix}$ Expanding the determinant: $\vec{n} = \mathbf{i}((1)(-1) - (0)(3)) - \mathbf{j}((3)(-1) - (0)(3)) + \mathbf{k}((3)(3) - (1)(3))$ $\vec{n} = \mathbf{i}(-1 - 0) - \mathbf{j}(-3 - 0) + \mathbf{k}(9 - 3)$ $\vec{n} = -1\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}$ So, the direction ratios of the normal vector are $(-1, 3, 6)$. We can use any scalar multiple of this vector as the normal vector. For instance, we can use $(1, -3, -6)$ by multiplying by $-1$.

Step 4: Write the equation of the plane Using the normal vector $\vec{n}=(A, B, C) = (1, -3, -6)$ and a point on the plane, say $P_1(-2, -2, 2)$, the equation of the plane is $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. $1(x - (-2)) - 3(y - (-2)) - 6(z - 2) = 0$ $1(x + 2) - 3(y + 2) - 6(z - 2) = 0$ $x + 2 - 3y - 6 - 6z + 12 = 0$ Combine the constant terms: $x - 3y - 6z + 8 = 0$ This is the general equation of the plane.

Step 5: Convert the general equation to intercept form The intercept form of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. To convert the general equation $x - 3y - 6z + 8 = 0$, we need to isolate the constant term on the right-hand side and make it equal to 1. First, move the constant term to the right side: $x - 3y - 6z = -8$ To achieve the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, we divide the entire equation by $-8$: $\frac{x}{-8} - \frac{3y}{-8} - \frac{6z}{-8} = \frac{-8}{-8}$ $\frac{x}{-8} + \frac{3y}{8} + \frac{6z}{8} = 1$ Now, rewrite each term to match the $\frac{x}{a}$ format: $\frac{x}{-8} + \frac{y}{8/3} + \frac{z}{8/6} = 1$ Simplify the denominators: $\frac{x}{-8} + \frac{y}{8/3} + \frac{z}{4/3} = 1$ From this equation, we can identify the intercepts:

• x-intercept ($a$) = $-8$
• y-intercept ($b$) = $8/3$
• z-intercept ($c$) = $4/3$

Step 6: Calculate the sum of the intercepts Sum of intercepts $= a + b + c$ Sum of intercepts $= -8 + \frac{8}{3} + \frac{4}{3}$ Sum of intercepts $= -8 + \frac{8+4}{3}$ Sum of intercepts $= -8 + \frac{12}{3}$ Sum of intercepts $= -8 + 4$ Sum of intercepts $= -4$

3. Common Mistakes & Tips

• Non-collinearity Check: Always ensure the three points are non-collinear before using the determinant method or cross product. If they are collinear, they define a line, not a unique plane.
• Determinant Expansion: Be very careful with the signs during determinant expansion. A single sign error can lead to an incorrect plane equation.
• Intercept Form Conversion: When converting $Ax + By + Cz + D = 0$ to intercept form, remember to move $D$ to the RHS as $-D$, so the equation becomes $Ax + By + Cz = -D$. Then divide by $-D$. A common mistake is to divide by $D$ when it should be $-D$, or to incorrectly move the constant term.

4. Summary

We first identified the three points lying on the plane. Then, we used two vectors formed by these points to calculate the normal vector to the plane via the cross product. With the normal vector and one of the points, we formulated the general equation of the plane. Finally, we converted this general equation into the intercept form to find the x, y, and z intercepts, and summed them up to get the final answer. The sum of the intercepts is -4.

5. Final Answer

The final answer is $\boxed{-4}$, which corresponds to option (B).