Key Concepts and Formulas

• Vector Representation: A vector connecting point $P$ to point $Q$ is given by $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$, where $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are the position vectors of $P$ and $Q$ respectively.
• Coplanarity of Vectors: Three vectors $\overrightarrow u, \overrightarrow v, \overrightarrow w$ are coplanar if and only if their scalar triple product (STP) is zero, i.e., $[\overrightarrow u \, \overrightarrow v \, \overrightarrow w] = 0$.
• Scalar Triple Product in Component Form: If $\overrightarrow u = u_1\overrightarrow a + u_2\overrightarrow b + u_3\overrightarrow c$, $\overrightarrow v = v_1\overrightarrow a + v_2\overrightarrow b + v_3\overrightarrow c$, and $\overrightarrow w = w_1\overrightarrow a + w_2\overrightarrow b + w_3\overrightarrow c$, where $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are non-coplanar, then their scalar triple product is given by the determinant: $[\overrightarrow u \, \overrightarrow v \, \overrightarrow w] = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}$

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Step-by-Step Solution

Step 1: Identify the Position Vectors of the Points We are given the position vectors of points $A, B, C,$ and $D$ with respect to the origin $O$.

• Position vector of $A$: $\overrightarrow{OA} = \overrightarrow a - \overrightarrow b + \overrightarrow c$
• Position vector of $B$: $\overrightarrow{OB} = \lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c$
• Position vector of $C$: $\overrightarrow{OC} = - \overrightarrow a + 2\overrightarrow b - 3\overrightarrow c$
• Position vector of $D$: $\overrightarrow{OD} = 2\overrightarrow a - 4\overrightarrow b + 6\overrightarrow c$

Step 2: Calculate the Vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ We need to find the vectors connecting point $A$ to points $B, C,$ and $D$. This is done by subtracting the position vector of the initial point from the position vector of the terminal point.

• Calculate $\overrightarrow{AB}$: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $\overrightarrow{AB} = (\lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c) - (\overrightarrow a - \overrightarrow b + \overrightarrow c)$ $\overrightarrow{AB} = (\lambda - 1)\overrightarrow a + (-3 - (-1))\overrightarrow b + (4 - 1)\overrightarrow c$ $\overrightarrow{AB} = (\lambda - 1)\overrightarrow a - 2\overrightarrow b + 3\overrightarrow c$ The components of $\overrightarrow{AB}$ with respect to the basis $\{\overrightarrow a, \overrightarrow b, \overrightarrow c\}$ are $(\lambda - 1, -2, 3)$.

• Calculate $\overrightarrow{AC}$: $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$ $\overrightarrow{AC} = (-\overrightarrow a + 2\overrightarrow b - 3\overrightarrow c) - (\overrightarrow a - \overrightarrow b + \overrightarrow c)$ $\overrightarrow{AC} = (-1 - 1)\overrightarrow a + (2 - (-1))\overrightarrow b + (-3 - 1)\overrightarrow c$ $\overrightarrow{AC} = -2\overrightarrow a + 3\overrightarrow b - 4\overrightarrow c$ The components of $\overrightarrow{AC}$ with respect to the basis $\{\overrightarrow a, \overrightarrow b, \overrightarrow c\}$ are $(-2, 3, -4)$.

• Calculate $\overrightarrow{AD}$: $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA}$ $\overrightarrow{AD} = (2\overrightarrow a - 4\overrightarrow b + 6\overrightarrow c) - (\overrightarrow a - \overrightarrow b + \overrightarrow c)$ $\overrightarrow{AD} = (2 - 1)\overrightarrow a + (-4 - (-1))\overrightarrow b + (6 - 1)\overrightarrow c$ $\overrightarrow{AD} = \overrightarrow a - 3\overrightarrow b + 5\overrightarrow c$ The components of $\overrightarrow{AD}$ with respect to the basis $\{\overrightarrow a, \overrightarrow b, \overrightarrow c\}$ are $(1, -3, 5)$.

Step 3: Apply the Coplanarity Condition We are given that the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ are coplanar. This means their scalar triple product is zero. Since $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are non-coplanar, they form a basis, and we can use the determinant of the components of these vectors.

The determinant formed by the components of $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ must be zero: $\begin{vmatrix} (\lambda - 1) & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$

Step 4: Solve the Determinant Equation for $\lambda$ We expand the determinant along the first row: $(\lambda - 1) \left| \begin{matrix} 3 & -4 \\ -3 & 5 \end{matrix} \right| - (-2) \left| \begin{matrix} -2 & -4 \\ 1 & 5 \end{matrix} \right| + 3 \left| \begin{matrix} -2 & 3 \\ 1 & -3 \end{matrix} \right| = 0$

Now, we evaluate the $2 \times 2$ determinants:

• $\left| \begin{matrix} 3 & -4 \\ -3 & 5 \end{matrix} \right| = (3)(5) - (-4)(-3) = 15 - 12 = 3$
• $\left| \begin{matrix} -2 & -4 \\ 1 & 5 \end{matrix} \right| = (-2)(5) - (-4)(1) = -10 - (-4) = -10 + 4 = -6$
• $\left| \begin{matrix} -2 & 3 \\ 1 & -3 \end{matrix} \right| = (-2)(-3) - (3)(1) = 6 - 3 = 3$

Substitute these values back into the expanded determinant equation: $(\lambda - 1)(3) + 2(-6) + 3(3) = 0$ $3(\lambda - 1) - 12 + 9 = 0$ $3(\lambda - 1) - 3 = 0$ Add 3 to both sides: $3(\lambda - 1) = 3$ Divide by 3: $\lambda - 1 = 1$ Add 1 to both sides: $\lambda = 2$

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Common Mistakes & Tips

• Sign Errors in Vector Subtraction: Be extremely careful when subtracting position vectors, especially when dealing with negative coefficients. For example, $2 - (-1) = 3$, not $1$.
• Determinant Calculation Errors: Double-check the arithmetic when calculating the $2 \times 2$ and $3 \times 3$ determinants. A single sign error can change the final answer.
• Understanding Basis Vectors: The problem states that $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are non-coplanar, which is essential for using the determinant method to find the scalar triple product.

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Summary

The problem requires us to find the value of $\lambda$ such that three vectors, formed by the position vectors of four points, are coplanar. We first calculated these vectors, $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$, by subtracting the position vector of point $A$ from the position vectors of points $B, C,$ and $D$ respectively. Since these vectors are coplanar, their scalar triple product must be zero. We expressed this condition as a determinant of the components of these vectors with respect to the non-coplanar basis vectors $\overrightarrow a, \overrightarrow b, \overrightarrow c$. Solving this determinant equation for $\lambda$ yielded the value 2.

The final answer is $\boxed{2}$.