Let $$\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$$, $$\vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k}$$ and a vector $$\vec{c}$$ be such that $$\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}$$. If $$\vec{a} \cdot \vec{c} = 13$$, then $$(24 - \vec{b} \cdot \vec{c})$$ is equal to _______

We are given
$$\vec a = 2\hat i - 3\hat j + 4\hat k,\; \vec b = 3\hat i + 4\hat j - 5\hat k,\; \vec c = x\hat i + y\hat j + z\hat k$$

The vector equation is
$$\vec a \times (\vec b + \vec c) + \vec b \times \vec c = \hat i + 8\hat j + 13\hat k \; -(1)$$

First expand the cross-product term:
$$\vec a \times (\vec b + \vec c)=\vec a \times \vec b+\vec a \times \vec c$$

Hence $$-(1)$$ becomes
$$\vec a \times \vec b + \vec a \times \vec c + \vec b \times \vec c = \hat i + 8\hat j + 13\hat k \; -(2)$$

Step 1 Compute $$\vec a \times \vec b$$ :

$$ \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 2 & -3 & 4\\[2pt] 3 & 4 & -5 \end{vmatrix} = \bigl((-3)(-5)-4\cdot4\bigr)\hat i - \bigl(2(-5)-4\cdot3\bigr)\hat j + \bigl(2\cdot4-(-3)\cdot3\bigr)\hat k$$ $$= (-1)\hat i + 22\hat j + 17\hat k$$ So $$\vec a \times \vec b = (-1,\,22,\,17).$$

Step 2 Write $$\vec a \times \vec c$$ :

$$ \vec a \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 2 & -3 & 4\\[2pt] x & y & z \end{vmatrix} = \bigl(-3z-4y\bigr)\hat i + \bigl(-2z+4x\bigr)\hat j + \bigl(2y+3x\bigr)\hat k.$$ Thus $$\vec a \times \vec c = (-3z-4y,\,-2z+4x,\,2y+3x).$$

Step 3 Write $$\vec b \times \vec c$$ :

$$ \vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 3 & 4 & -5\\[2pt] x & y & z \end{vmatrix} = (4z+5y)\hat i + \bigl(-(3z+5x)\bigr)\hat j + (3y-4x)\hat k.$$ Hence $$\vec b \times \vec c = (4z+5y,\,-3z-5x,\,3y-4x).$$

Step 4 Add $$\vec a \times \vec c$$ and $$\vec b \times \vec c$$ :

$$\vec a \times \vec c + \vec b \times \vec c = (z+y,\,-5z-x,\,5y-x).$$

Step 5 Insert everything into $$-(2)$$ :

$$(\underbrace{-1,\,22,\,17}_{\vec a \times \vec b}) + (z+y,\,-5z-x,\,5y-x) = (1,\,8,\,13).$$

Equate components:

$$\begin{aligned} z + y &= 2 \quad -(3)\\[2pt] x + 5z &= 14 \quad -(4)\\[2pt] x - 5y &= 4 \quad -(5) \end{aligned}$$

Step 6 Use the additional condition
$$\vec a \cdot \vec c = 13.$$

Since $$\vec a \cdot \vec c = 2x - 3y + 4z,$$ we have

$$2x - 3y + 4z = 13 \quad -(6)$$

Step 7 Solve the system.

From $$(3):\; y = 2 - z.$$

Substitute $$y$$ in $$(5):$$ $$x - 5(2 - z) = 4 \;\Longrightarrow\; x + 5z = 14,$$ which is the same as $$(4)$$, so the equations are consistent.

Now use $$(4)$$ and $$(6):$$
From $$(4):\; x = 14 - 5z.$

Put this in $$(6):$$ $$2(14 - 5z) - 3(2 - z) + 4z = 13$$ $$28 - 10z - 6 + 3z + 4z = 13$$ $$22 - 3z = 13 \;\Longrightarrow\; 3z = 9 \;\Longrightarrow\; z = 3.$$

Now
$$y = 2 - z = 2 - 3 = -1,$$ $$x = 14 - 5z = 14 - 15 = -1.$$

Therefore $$$$\vec$$ c = -$$\hat$$ i - $$\hat$$ j + 3$$\hat$$ k.$$

Step 8 Find $$$$\vec$$ b $$\cdot$$ $$\vec$$ c$$ :

$$$$\vec$$ b $$\cdot$$ $$\vec$$ c = 3x + 4y - 5z = 3(-1) + 4(-1) - 5(3) = -3 - 4 - 15 = -22.$$

Step 9 Compute the required expression:

$$24 - $$\vec$$ b $$\cdot$$ $$\vec$$ c = 24 - (-22) = 24 + 22 = 46.$$

Hence the value of $$\bigl(24 - $$\vec$$ b $$\cdot$$ $$\vec$$ c\bigr)$$ is $$46$$.