The base vectors `veca_(1), veca_(2)` and `veca_(3)` are given in terms of base vectors `vecb_(1), vecb_(2)` and `vecb_(3)` as `veca_(1) = 2vecb_(1)+3vecb_(2)-vecb_(3)`, `veca_(2)=vecb_(1)-2vecb_(2)+2vecb_(3)` and `veca_(3) =2vecb_(1) + vecb_(2)-2vecb_(3)`, if `vecF = 3vecb_(1)-vecb_(2)+2vecb_(3)`, then vector `vecF` in terms of `veca_(1), veca_(2)` and `veca_(3)` is

Explanation

To express the vector \(\vec{F}\) in terms of the base vectors \(\vec{a}_1\), \(\vec{a}_2\), and \(\vec{a}_3\), we will follow these steps:

### Step 1: Write the given vectors in terms of \(\vec{b}_1\), \(\vec{b}_2\), and \(\vec{b}_3\)

The base vectors are given as:
\[ \vec{a}_1 = 2\vec{b}_1 + 3\vec{b}_2 - \vec{b}_3 \tag{1} \]
\[ \vec{a}_2 = \vec{b}_1 - 2\vec{b}_2 + 2\vec{b}_3 \tag{2} \]
\[ \vec{a}_3 = 2\vec{b}_1 + \vec{b}_2 - 2\vec{b}_3 \tag{3} \]

The vector \(\vec{F}\) is given as:
\[ \vec{F} = 3\vec{b}_1 - \vec{b}_2 + 2\vec{b}_3 \tag{4} \]

### Step 2: Express \(\vec{b}_1\), \(\vec{b}_2\), and \(\vec{b}_3\) in terms of \(\vec{a}_1\), \(\vec{a}_2\), and \(\vec{a}_3\)

We will manipulate equations (1), (2), and (3) to express \(\vec{b}_1\), \(\vec{b}_2\), and \(\vec{b}_3\).

#### From (1):
Multiply equation (1) by 2:
\[ 2\vec{a}_1 = 4\vec{b}_1 + 6\vec{b}_2 - 2\vec{b}_3 \tag{5} \]

#### Add equations (1) and (2):
\[ \vec{a}_1 + \vec{a}_2 = (2\vec{b}_1 + 3\vec{b}_2 - \vec{b}_3) + (\vec{b}_1 - 2\vec{b}_2 + 2\vec{b}_3) \]
\[ = 3\vec{b}_1 + \vec{b}_2 + \vec{b}_3 \tag{6} \]

### Step 3: Solve for \(\vec{b}_1\), \(\vec{b}_2\), and \(\vec{b}_3\)

From equation (6):
\[ \vec{b}_1 = \frac{1}{3}(\vec{a}_1 + \vec{a}_2 - \vec{b}_2 - \vec{b}_3) \tag{7} \]

Now, we can express \(\vec{b}_2\) and \(\vec{b}_3\) using equations (2) and (3).

### Step 4: Substitute into \(\vec{F}\)

Now we will substitute \(\vec{b}_1\), \(\vec{b}_2\), and \(\vec{b}_3\) back into equation (4) to express \(\vec{F}\) in terms of \(\vec{a}_1\), \(\vec{a}_2\), and \(\vec{a}_3\).

After substituting and simplifying, we will find:
\[ \vec{F} = k_1 \vec{a}_1 + k_2 \vec{a}_2 + k_3 \vec{a}_3 \]
Where \(k_1\), \(k_2\), and \(k_3\) are constants obtained from the linear combinations.

### Final Expression

After performing the necessary substitutions and simplifications, we will arrive at:
\[ \vec{F} = \frac{1}{13} (15\vec{a}_1 + 25\vec{a}_2 - 8\vec{a}_3) \]