Find `vec(a).(vec(b)xx vec(c ))` if :
`vec(a)=2hat(i)+hat(j)+3hat(k), vec(b)=-hat(i)+2hat(j)+hat(k)` and `vec(c )=3hat(i)+hat(j)+2hat(k)`.

Answer

To find the scalar triple product \(\vec{a} \cdot (\vec{b} \times \vec{c})\), we can use the determinant method. The scalar triple product can be represented as the determinant of a matrix formed by the components of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step 1: Write down the vectors
Given:
\[ \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \]
\[ \vec{b} = -\hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{c} = 3\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Set up the determinant
The scalar triple product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) can be calculated using the determinant:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 2 & 1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can calculate the determinant using the formula:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this to our determinant:
\[ = 2 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} -1 & 1 \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2) - (1 \cdot 1) = 4 - 1 = 3\)

2. \(\begin{vmatrix} -1 & 1 \\ 3 & 2 \end{vmatrix} = (-1 \cdot 2) - (1 \cdot 3) = -2 - 3 = -5\)

3. \(\begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = (-1 \cdot 1) - (2 \cdot 3) = -1 - 6 = -7\)

Now substituting back:
\[ = 2(3) - 1(-5) + 3(-7) \]
\[ = 6 + 5 - 21 \]
\[ = 11 - 21 = -10 \]

### Final Answer
Thus, the value of \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is \(-10\).

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