Key Concepts and Formulas

1. Vector Operations: Understanding of vector addition, subtraction, scalar multiplication, dot product, and cross product.
   • Dot Product: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$ (scalar).
   • Cross Product: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$ (vector).
2. Vector Component Representation: A vector $\vec{c}$ can be represented as $\vec{c} = c_x \hat{i} + c_y \hat{j} + c_z \hat{k}$.
3. Magnitude of a Vector: For a vector $\vec{c} = c_x \hat{i} + c_y \hat{j} + c_z \hat{k}$, its magnitude squared is $|\vec{c}|^2 = c_x^2 + c_y^2 + c_z^2$.
4. Equality of Vectors: Two vectors are equal if and only if their corresponding components are equal.

Step-by-Step Solution

Step 1: Define the unknown vector and calculate preliminary vector expressions. Let the unknown vector be $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. We are given: $\vec{a} = 3\hat{i} + 2\hat{j} + \hat{k}$ $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$

Calculate $\vec{a}+\vec{b}$: $\vec{a}+\vec{b} = (3+2)\hat{i} + (2-1)\hat{j} + (1+3)\hat{k} = 5\hat{i} + \hat{j} + 4\hat{k}$ Calculate $\vec{a}-\vec{b}$: $\vec{a}-\vec{b} = (3-2)\hat{i} + (2-(-1))\hat{j} + (1-3)\hat{k} = \hat{i} + 3\hat{j} - 2\hat{k}$ Calculate $\vec{a} \times \vec{b}$: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \hat{i}(6 - (-1)) - \hat{j}(9 - 2) + \hat{k}(-3 - 4) = 7\hat{i} - 7\hat{j} - 7\hat{k}$

Step 2: Use the first vector equation to form scalar equations. The first equation is $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$.

Calculate the LHS: $(\vec{a}+\vec{b}) \times \vec{c}$ $(5\hat{i} + \hat{j} + 4\hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix}$ $= \hat{i}(z - 4y) - \hat{j}(5z - 4x) + \hat{k}(5y - x) = (z-4y)\hat{i} + (4x-5z)\hat{j} + (5y-x)\hat{k}$ Calculate the RHS: $2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ $2(7\hat{i} - 7\hat{j} - 7\hat{k}) + 24\hat{j} - 6\hat{k} = 14\hat{i} - 14\hat{j} - 14\hat{k} + 24\hat{j} - 6\hat{k}$ $= 14\hat{i} + 10\hat{j} - 20\hat{k}$ Equating the components of LHS and RHS:

1. $z - 4y = 14$
2. $4x - 5z = 10$
3. $5y - x = -20$

Step 3: Use the second vector equation to form another scalar equation. The second equation is $(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$.

Calculate the vector term: $\vec{a}-\vec{b}+\hat{i}$ $(\hat{i} + 3\hat{j} - 2\hat{k}) + \hat{i} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ Calculate the dot product with $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$: $(2\hat{i} + 3\hat{j} - 2\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = -3$ $2x + 3y - 2z = -3$ This gives us the fourth scalar equation: 4. $2x + 3y - 2z = -3$

Step 4: Solve the system of linear scalar equations. We have the system:

1. $-4y + z = 14 \implies z = 4y + 14$
2. $4x - 5z = 10$
3. $-x + 5y = -20 \implies x = 5y + 20$
4. $2x + 3y - 2z = -3$

Substitute the expressions for $x$ and $z$ from (3) and (1) into (4): $2(5y+20) + 3y - 2(4y+14) = -3$ $10y + 40 + 3y - 8y - 28 = -3$ Combine terms: $(10y + 3y - 8y) + (40 - 28) = -3$ $5y + 12 = -3$ $5y = -15$ $y = -3$ Now, find $x$ and $z$ using the expressions derived earlier: $x = 5y + 20 = 5(-3) + 20 = -15 + 20 = 5$ $z = 4y + 14 = 4(-3) + 14 = -12 + 14 = 2$ So, $\vec{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$.

Step 5: Calculate $|\vec{c}|^2$. Using the components $x=5$, $y=-3$, and $z=2$: $|\vec{c}|^2 = x^2 + y^2 + z^2$ $|\vec{c}|^2 = (5)^2 + (-3)^2 + (2)^2$ $|\vec{c}|^2 = 25 + 9 + 4$ $|\vec{c}|^2 = 38$

Common Mistakes & Tips

• Cross Product Calculation Errors: Be meticulous with the determinant expansion and sign conventions for the cross product. A common mistake is miscalculating the $\hat{j}$ component.
• Algebraic Errors in Solving System: Solving systems of linear equations can lead to arithmetic errors. Double-check substitutions and calculations.
• Component-wise Operations: Ensure all vector operations are performed correctly on their respective components.

Summary The problem involves finding an unknown vector $\vec{c}$ given two vector equations. The strategy is to express $\vec{c}$ in component form, $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$, and then translate the vector equations into a system of linear scalar equations for $x, y, z$. After calculating the necessary preliminary vector operations (addition, subtraction, cross product), we set up the equations. The first vector equation yielded three scalar equations, and the second vector equation provided a fourth. Solving this system of linear equations allowed us to determine the components of $\vec{c}$. Finally, the magnitude squared of $\vec{c}$ was calculated using its components.

The final answer is $\boxed{38}$.