Key Concepts and Formulas

• Coplanarity of vectors: Three vectors $\overrightarrow u, \overrightarrow v, \overrightarrow w$ are coplanar if their scalar triple product is zero, i.e., $[\overrightarrow u, \overrightarrow v, \overrightarrow w] = \overrightarrow u \cdot (\overrightarrow v \times \overrightarrow w) = 0$. Alternatively, a vector $\overrightarrow c$ is coplanar with $\overrightarrow a$ and $\overrightarrow b$ if $\overrightarrow c$ can be expressed as a linear combination of $\overrightarrow a$ and $\overrightarrow b$, i.e., $\overrightarrow c = p\overrightarrow a + q\overrightarrow b$ for some scalars $p, q$.
• Scalar Triple Product (STP): For vectors $\overrightarrow u = u_x\widehat i + u_y\widehat j + u_z\widehat k$, $\overrightarrow v = v_x\widehat i + v_y\widehat j + v_z\widehat k$, and $\overrightarrow w = w_x\widehat i + w_y\widehat j + w_z\widehat k$, the STP is given by $[\overrightarrow u, \overrightarrow v, \overrightarrow w] = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$.
• Cross Product: The cross product of two vectors $\overrightarrow u$ and $\overrightarrow v$ is given by $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$.
• Dot Product: The dot product of two vectors $\overrightarrow u$ and $\overrightarrow v$ is given by $\overrightarrow u \cdot \overrightarrow v = u_x v_x + u_y v_y + u_z v_z$. The dot product with $\widehat k$ isolates the $z$-component of a vector: $(\overrightarrow V) \cdot \widehat k = V_z$.
• Magnitude of a vector: For a vector $\overrightarrow c = c_x\widehat i + c_y\widehat j + c_z\widehat k$, its squared magnitude is $|\overrightarrow c|^2 = c_x^2 + c_y^2 + c_z^2$.

Step-by-Step Solution

We are given $\overrightarrow a = - \widehat i + \widehat j + \widehat k$ and $\overrightarrow b = 2\widehat i + \widehat j - \widehat k$. Let $\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$.

Step 1: Pre-compute auxiliary vectors. To simplify the subsequent calculations, we compute the vectors $\overrightarrow a + \overrightarrow b$ and $\overrightarrow a - \overrightarrow b$. $\overrightarrow a + \overrightarrow b = (-1 + 2)\widehat i + (1 + 1)\widehat j + (1 - 1)\widehat k = \widehat i + 2\widehat j$ $\overrightarrow a - \overrightarrow b = (-1 - 2)\widehat i + (1 - 1)\widehat j + (1 - (-1))\widehat k = -3\widehat i + 2\widehat k$

Step 2: Compute the cross product $\overrightarrow a \times \overrightarrow b$. This vector is crucial for the coplanarity condition and for the second given condition. $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ -1 & 1 & 1 \\ 2 & 1 & -1 \end{vmatrix} = \widehat i((1)(-1) - (1)(1)) - \widehat j((-1)(-1) - (1)(2)) + \widehat k((-1)(1) - (1)(2))$ $= \widehat i(-1 - 1) - \widehat j(1 - 2) + \widehat k(-1 - 2) = -2\widehat i + \widehat j - 3\widehat k$

Step 3: Apply the coplanarity condition. The condition that $\overrightarrow c$ is coplanar with $\overrightarrow a$ and $\overrightarrow b$ means that $\overrightarrow c \cdot (\overrightarrow a \times \overrightarrow b) = 0$. Substituting the components of $\overrightarrow c$ and the computed $\overrightarrow a \times \overrightarrow b$: $(x\widehat i + y\widehat j + z\widehat k) \cdot (-2\widehat i + \widehat j - 3\widehat k) = 0$ $-2x + y - 3z = 0 \quad (*)$

Step 4: Apply the second given condition. The second condition is $\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$. First, we compute the cross product $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)$: $\left( {\widehat i + 2\widehat j} \right) \times \left( {-2\widehat i + \widehat j - 3\widehat k} \right) = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 2 & 0 \\ -2 & 1 & -3 \end{vmatrix}$ $= \widehat i((2)(-3) - (0)(1)) - \widehat j((1)(-3) - (0)(-2)) + \widehat k((1)(1) - (2)(-2))$ $= \widehat i(-6 - 0) - \widehat j(-3 - 0) + \widehat k(1 + 4) = -6\widehat i + 3\widehat j + 5\widehat k$ Now, we take the dot product with $\overrightarrow c$: $(x\widehat i + y\widehat j + z\widehat k) \cdot (-6\widehat i + 3\widehat j + 5\widehat k) = -42$ $-6x + 3y + 5z = -42 \quad (**)$

Step 5: Apply the third given condition. The third condition is $\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$. First, we compute the cross product $\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)$: $\overrightarrow c \times \left( {-3\widehat i + 2\widehat k} \right) = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ x & y & z \\ -3 & 0 & 2 \end{vmatrix}$ $= \widehat i((y)(2) - (z)(0)) - \widehat j((x)(2) - (z)(-3)) + \widehat k((x)(0) - (y)(-3))$ $= 2y\widehat i - (2x + 3z)\widehat j + 3y\widehat k$ Now, we take the dot product with $\widehat k$: $(2y\widehat i - (2x + 3z)\widehat j + 3y\widehat k) \cdot \widehat k = 3$ The dot product with $\widehat k$ extracts the $z$-component of the vector: $3y = 3$ $y = 1$

Step 6: Solve the system of linear equations. We have the following system of equations:

1. $-2x + y - 3z = 0$
2. $-6x + 3y + 5z = -42$
3. $y = 1$

Substitute $y=1$ into equation (1): $-2x + 1 - 3z = 0 \implies -2x - 3z = -1 \implies 2x + 3z = 1 \quad (***)$ Substitute $y=1$ into equation (2): $-6x + 3(1) + 5z = -42 \implies -6x + 3 + 5z = -42 \implies -6x + 5z = -45 \quad (****)$ Now we solve the system formed by (*) and (**): $2x + 3z = 1$ $-6x + 5z = -45$ Multiply the first equation by 3: $6x + 9z = 3$ Add this to the second equation: $(6x + 9z) + (-6x + 5z) = 3 + (-45)$ $14z = -42$ $z = \frac{-42}{14} = -3$ Substitute $z=-3$ back into $2x + 3z = 1$: $2x + 3(-3) = 1$ $2x - 9 = 1$ $2x = 10$ $x = 5$ So, the components of $\overrightarrow c$ are $x=5$, $y=1$, and $z=-3$. Thus, $\overrightarrow c = 5\widehat i + \widehat j - 3\widehat k$.

Step 7: Calculate $|\overrightarrow c|^2$. $|\overrightarrow c|^2 = x^2 + y^2 + z^2$ $|\overrightarrow c|^2 = (5)^2 + (1)^2 + (-3)^2$ $|\overrightarrow c|^2 = 25 + 1 + 9$ $|\overrightarrow c|^2 = 35$

Common Mistakes & Tips

• Sign errors in cross products: Be extremely careful with the signs, especially for the $\widehat j$ component when calculating determinants for cross products.
• Algebraic manipulation: Double-check each step when solving the system of linear equations to avoid calculation mistakes.
• Coplanarity interpretation: Remember that coplanarity implies the scalar triple product is zero, which translates to a linear relationship between the vector components.

Summary

We systematically used the given conditions to form a system of linear equations for the components of vector $\overrightarrow c$. The coplanarity condition provided the first equation. The second condition involved calculating a vector triple product and then a dot product, yielding a second linear equation. The third condition simplified to directly giving the $y$-component of $\overrightarrow c$. Solving the resulting system of linear equations allowed us to find the components of $\overrightarrow c$ and subsequently its squared magnitude.

The final answer is $\boxed{35}$.