Write the projection of `vecb + vecc` on `veca` where
`veca = 2 hati- 2 hatj +hatk, vecb = hati+ 2hatj- 2hatk and vecc = 2 hati- hatj+ 4 hatk`

Answer

To find the projection of the vector \(\vec{b} + \vec{c}\) on the vector \(\vec{a}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 2\hat{i} - 2\hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \]
\[ \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \]

### Step 2: Calculate \(\vec{b} + \vec{c}\)
We need to add the vectors \(\vec{b}\) and \(\vec{c}\):
\[ \vec{b} + \vec{c} = (\hat{i} + 2\hat{j} - 2\hat{k}) + (2\hat{i} - \hat{j} + 4\hat{k}) \]
Combining like terms:
\[ = (1 + 2)\hat{i} + (2 - 1)\hat{j} + (-2 + 4)\hat{k} \]
\[ = 3\hat{i} + 1\hat{j} + 2\hat{k} \]
Thus,
\[ \vec{b} + \vec{c} = 3\hat{i} + \hat{j} + 2\hat{k} \]

### Step 3: Calculate the dot product \((\vec{b} + \vec{c}) \cdot \vec{a}\)
Now we need to find the dot product of \(\vec{b} + \vec{c}\) with \(\vec{a}\):
\[ \vec{a} \cdot (\vec{b} + \vec{c}) = (2\hat{i} - 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j} + 2\hat{k}) \]
Calculating the dot product:
\[ = 2 \cdot 3 + (-2) \cdot 1 + 1 \cdot 2 \]
\[ = 6 - 2 + 2 \]
\[ = 6 \]

### Step 4: Calculate the magnitude of \(\vec{a}\)
Next, we calculate the magnitude of \(\vec{a}\):
\[ |\vec{a}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} \]
\[ = \sqrt{4 + 4 + 1} \]
\[ = \sqrt{9} = 3 \]

### Step 5: Calculate the projection
Finally, we can find the projection of \(\vec{b} + \vec{c}\) on \(\vec{a}\):
\[ \text{Projection of } (\vec{b} + \vec{c}) \text{ on } \vec{a} = \frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a}|} \]
\[ = \frac{6}{3} = 2 \]

### Final Answer
Thus, the projection of \(\vec{b} + \vec{c}\) on \(\vec{a}\) is:
\[ \boxed{2} \]