1. Key Concepts and Formulas

   • Distributive Property of Cross Product: $\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$
   • Cyclic Property of Cross Product: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$
   • Scalar Triple Product: $[\vec{A} \ \vec{B} \ \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C})$. If any two vectors are identical, the scalar triple product is zero.
   • Vector Component Representation: A vector $\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$ can be represented by its components.
   • Dot Product: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
   • Cross Product: $\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$.

2. Step-by-Step Solution

   Step 1: Simplify the given vector equation. We are given the equation $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. Apply the distributive property of the cross product to the first term: $\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \hat{i}+8 \hat{j}+13 \hat{k}$ Rearrange the terms and use the property $\vec{a} \times \vec{c} = -(\vec{c} \times \vec{a})$: $\vec{a} \times \vec{b} + (\vec{a} + \vec{b}) \times \vec{c} = \hat{i}+8 \hat{j}+13 \hat{k}$ This form is useful because it isolates the unknown vector $\vec{c}$ in a cross product with a known vector sum.

   Step 2: Calculate the known vectors and their cross product. Given $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$. First, calculate $\vec{a} \times \vec{b}$: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 3 & 4 & -5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(-10-12) + \hat{k}(8-(-9)) = -\hat{i} + 22\hat{j} + 17\hat{k}$ Next, calculate the sum $\vec{a}+\vec{b}$: $\vec{a}+\vec{b} = (2+3)\hat{i} + (-3+4)\hat{j} + (4-5)\hat{k} = 5\hat{i} + \hat{j} - \hat{k}$ Let $\vec{d} = \vec{a}+\vec{b} = 5\hat{i} + \hat{j} - \hat{k}$.

   Step 3: Isolate the term involving $\vec{c}$. Substitute the calculated $\vec{a} \times \vec{b}$ back into the simplified equation from Step 1: $(-\hat{i} + 22\hat{j} + 17\hat{k}) + (5\hat{i} + \hat{j} - \hat{k}) \times \vec{c} = \hat{i}+8 \hat{j}+13 \hat{k}$ Rearrange to solve for $(\vec{a}+\vec{b}) \times \vec{c}$: $(5\hat{i} + \hat{j} - \hat{k}) \times \vec{c} = (\hat{i}+8 \hat{j}+13 \hat{k}) - (-\hat{i} + 22\hat{j} + 17\hat{k})$ $(5\hat{i} + \hat{j} - \hat{k}) \times \vec{c} = (1 - (-1))\hat{i} + (8 - 22)\hat{j} + (13 - 17)\hat{k}$ $(5\hat{i} + \hat{j} - \hat{k}) \times \vec{c} = 2\hat{i} - 14\hat{j} - 4\hat{k}$ Let $\vec{X} = 2\hat{i} - 14\hat{j} - 4\hat{k}$. So, we have $\vec{d} \times \vec{c} = \vec{X}$.

   Step 4: Form a system of linear equations for the components of $\vec{c}$. Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. We have two conditions involving $\vec{c}$: (i) $\vec{a} \cdot \vec{c} = 13$ (ii) $(\vec{a}+\vec{b}) \times \vec{c} = 2\hat{i} - 14\hat{j} - 4\hat{k}$

   From condition (i): $(2\hat{i}-3\hat{j}+4\hat{k}) \cdot (x\hat{i}+y\hat{j}+z\hat{k}) = 13$ $2x - 3y + 4z = 13 \quad \text{(Equation 1)}$

   From condition (ii), calculate the cross product $(\vec{a}+\vec{b}) \times \vec{c}$: $(5\hat{i} + \hat{j} - \hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & -1 \\ x & y & z \end{vmatrix}$ $= \hat{i}(z - (-y)) - \hat{j}(5z - (-x)) + \hat{k}(5y - x)$ $= (z+y)\hat{i} - (5z+x)\hat{j} + (5y-x)\hat{k}$ Equating this to $2\hat{i} - 14\hat{j} - 4\hat{k}$: $z+y = 2 \quad \text{(Equation 2)}$ $-(5z+x) = -14 \implies x+5z = 14 \quad \text{(Equation 3)}$ $5y-x = -4 \quad \text{(Equation 4)}$ Note that Equation 4 is dependent on Equations 2 and 3. We can verify this by expressing $x$ and $y$ from (2) and (3) in terms of $z$ and substituting into (4). From (2), $y=2-z$. From (3), $x=14-5z$. Substituting into (4): $5(2-z) - (14-5z) = 10-5z-14+5z = -4$, which is consistent. Thus, we have three independent equations (Equation 1, Equation 2, Equation 3).

   Step 5: Solve the system of linear equations. We have:

   1. $2x - 3y + 4z = 13$
   2. $y+z = 2 \implies y = 2-z$
   3. $x+5z = 14 \implies x = 14-5z$

   Substitute the expressions for $x$ and $y$ from (2) and (3) into Equation 1: $2(14-5z) - 3(2-z) + 4z = 13$ $28 - 10z - 6 + 3z + 4z = 13$ $22 - 3z = 13$ $3z = 22 - 13$ $3z = 9$ $z = 3$ Now find $x$ and $y$: $y = 2 - z = 2 - 3 = -1$ $x = 14 - 5z = 14 - 5(3) = 14 - 15 = -1$ So, the vector $\vec{c}$ is $\vec{c} = -1\hat{i} - 1\hat{j} + 3\hat{k}$.

   Step 6: Calculate $\vec{b} \cdot \vec{c}$. We have $\vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=-\hat{i}-\hat{j}+3\hat{k}$. $\vec{b} \cdot \vec{c} = (3)(-1) + (4)(-1) + (-5)(3)$ $= -3 - 4 - 15$ $= -22$

   Step 7: Calculate the final expression $24-\vec{b} \cdot \vec{c}$. $24 - \vec{b} \cdot \vec{c} = 24 - (-22)$ $= 24 + 22$ $= 46$

3. Common Mistakes & Tips

   • Sign Errors in Cross Products: Be meticulous with the signs when calculating the determinant for the cross product. The middle term for the $\hat{j}$ component is typically subtracted.
   • Algebraic Manipulation: Solving systems of linear equations can lead to errors. Double-check substitutions and arithmetic.
   • Vector Properties: Ensure correct application of distributive and cyclic properties of vector operations.

4. Summary The problem involves simplifying a vector equation using the distributive property of the cross product. This led to an equation of the form $(\vec{a}+\vec{b}) \times \vec{c} = \vec{X}$, where $\vec{a}+\vec{b}$ and $\vec{X}$ are known vectors. Combined with the given condition $\vec{a} \cdot \vec{c} = 13$, we formed a system of three linear equations for the components of $\vec{c}$. Solving this system yielded $\vec{c} = -\hat{i} - \hat{j} + 3\hat{k}$. Finally, we computed $\vec{b} \cdot \vec{c} = -22$, and the required expression $24 - \vec{b} \cdot \vec{c}$ evaluates to $24 - (-22) = 46$.

The final answer is $\boxed{46}$.