Key Concepts and Formulas

• Coplanarity of Four Points: Four points A, B, C, and D are coplanar if the vectors formed by taking one point as the origin and the other three points as endpoints are coplanar. This means the scalar triple product of these three vectors is zero: $[\vec{AB} \ \vec{AC} \ \vec{AD}] = 0$.
• Scalar Triple Product: For vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, their scalar triple product is given by the determinant of their components: $[\vec{a} \ \vec{b} \ \vec{c}] = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$
• Vector Between Two Points: If the position vectors of points P and Q are $\vec{OP}$ and $\vec{OQ}$ respectively, then the vector $\vec{PQ} = \vec{OQ} - \vec{OP}$.

Step-by-Step Solution

Step 1: Define the position vectors and form the relative vectors. We are given the position vectors of points A, B, C, and D: $\vec{OA} = 5 \hat{i}+5 \hat{j}+2 \lambda \hat{k}$ $\vec{OB} = \hat{i}+2 \hat{j}+3 \hat{k}$ $\vec{OC} = -2 \hat{i}+\lambda \hat{j}+4 \hat{k}$ $\vec{OD} = -\hat{i}+5 \hat{j}+6 \hat{k}$

To check for coplanarity, we form three vectors originating from a common point, say A. These vectors are $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$.

Step 2: Calculate the components of the relative vectors. Using the formula $\vec{PQ} = \vec{OQ} - \vec{OP}$:

• Calculate $\vec{AB}$: $\vec{AB} = \vec{OB} - \vec{OA} = (\hat{i}+2 \hat{j}+3 \hat{k}) - (5 \hat{i}+5 \hat{j}+2 \lambda \hat{k})$ $\vec{AB} = (1-5)\hat{i} + (2-5)\hat{j} + (3-2\lambda)\hat{k} = -4 \hat{i} - 3 \hat{j} + (3-2\lambda) \hat{k}$

• Calculate $\vec{AC}$: $\vec{AC} = \vec{OC} - \vec{OA} = (-2 \hat{i}+\lambda \hat{j}+4 \hat{k}) - (5 \hat{i}+5 \hat{j}+2 \lambda \hat{k})$ $\vec{AC} = (-2-5)\hat{i} + (\lambda-5)\hat{j} + (4-2\lambda)\hat{k} = -7 \hat{i} + (\lambda-5) \hat{j} + (4-2\lambda) \hat{k}$

• Calculate $\vec{AD}$: $\vec{AD} = \vec{OD} - \vec{OA} = (-\hat{i}+5 \hat{j}+6 \hat{k}) - (5 \hat{i}+5 \hat{j}+2 \lambda \hat{k})$ $\vec{AD} = (-1-5)\hat{i} + (5-5)\hat{j} + (6-2\lambda)\hat{k} = -6 \hat{i} + 0 \hat{j} + (6-2\lambda) \hat{k}$

Step 3: Apply the coplanarity condition. For points A, B, C, and D to be coplanar, the scalar triple product of $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ must be zero. $[\vec{AB} \ \vec{AC} \ \vec{AD}] = \left| \begin{array}{ccc} -4 & -3 & (3-2 \lambda) \\ -7 & (\lambda-5) & (4-2 \lambda) \\ -6 & 0 & (6-2 \lambda) \end{array} \right| = 0$

Step 4: Expand and solve the determinant. We expand the determinant. Expanding along the third row simplifies calculations due to the presence of a zero term. The determinant is: $-6 \left| \begin{array}{cc} -3 & (3-2 \lambda) \\ (\lambda-5) & (4-2 \lambda) \end{array} \right| - 0 \left| \begin{array}{cc} -4 & (3-2 \lambda) \\ -7 & (4-2 \lambda) \end{array} \right| + (6-2 \lambda) \left| \begin{array}{cc} -4 & -3 \\ -7 & (\lambda-5) \end{array} \right| = 0$

Let's calculate the determinants of the $2 \times 2$ matrices:

• First $2 \times 2$ determinant: $(-3)(4-2\lambda) - (3-2\lambda)(\lambda-5)$ $= (-12 + 6\lambda) - (3\lambda - 15 - 2\lambda^2 + 10\lambda)$ $= -12 + 6\lambda - (13\lambda - 15 - 2\lambda^2)$ $= -12 + 6\lambda - 13\lambda + 15 + 2\lambda^2$ $= 2\lambda^2 - 7\lambda + 3$

• Second $2 \times 2$ determinant: $(-4)(\lambda-5) - (-3)(-7)$ $= -4\lambda + 20 - 21$ $= -4\lambda - 1$

Now substitute these back into the expanded determinant equation: $-6(2\lambda^2 - 7\lambda + 3) + (6-2\lambda)(-4\lambda - 1) = 0$ Expand and simplify: $(-12\lambda^2 + 42\lambda - 18) + (-24\lambda - 6 + 8\lambda^2 + 2\lambda) = 0$ $-12\lambda^2 + 42\lambda - 18 + 8\lambda^2 - 22\lambda - 6 = 0$ Combine like terms: $(-12\lambda^2 + 8\lambda^2) + (42\lambda - 22\lambda) + (-18 - 6) = 0$ $-4\lambda^2 + 20\lambda - 24 = 0$ Divide the equation by $-4$: $\lambda^2 - 5\lambda + 6 = 0$

Step 5: Solve the quadratic equation for $\lambda$. Factor the quadratic equation: $(\lambda - 2)(\lambda - 3) = 0$ This yields two possible values for $\lambda$: $\lambda = 2 \quad \text{or} \quad \lambda = 3$

Step 6: Determine the set S and calculate the required sum. The set $S$ consists of all values of $\lambda$ for which the points are coplanar. Therefore, $S = \{2, 3\}$.

We need to calculate $\sum_{\lambda \in S}(\lambda+2)^2$. This means we sum the expression $(\lambda+2)^2$ for each element in $S$.

• For $\lambda = 2$: $(\lambda+2)^2 = (2+2)^2 = 4^2 = 16$

• For $\lambda = 3$: $(\lambda+2)^2 = (3+2)^2 = 5^2 = 25$

The sum is: $\sum_{\lambda \in S}(\lambda+2)^2 = 16 + 25 = 41$

Common Mistakes & Tips

• Vector Subtraction Errors: Be meticulous when subtracting position vectors. Ensure that the components are subtracted in the correct order ($\vec{OQ} - \vec{OP}$).
• Determinant Expansion Mistakes: Double-check the signs and calculations during determinant expansion, especially when dealing with negative numbers or variables. Expanding along a row or column with zeros can significantly reduce errors.
• Quadratic Equation Factoring: If factoring is difficult, use the quadratic formula ($\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$) to find the roots of the quadratic equation.

Summary The problem requires us to find the values of $\lambda$ for which four given points are coplanar. This condition is met when the scalar triple product of the vectors formed by these points is zero. We formed three vectors $\vec{AB}, \vec{AC}, \vec{AD}$, set their scalar triple product (represented by a determinant) to zero, and solved the resulting quadratic equation for $\lambda$. The set $S$ of these $\lambda$ values was found to be $\{2, 3\}$. Finally, we computed the sum of $(\lambda+2)^2$ for each $\lambda$ in $S$, yielding the result 41.

The final answer is \boxed{41}.