Find the value of `'m'` for which the line `vecr= (hati+2hatj-hatk)+lambda(2hati+hatj+hatk)` is parallel to the plane `vecr.(3hati-2hatj+mhatk) = 5`.

Answer

To find the value of \( m \) for which the line

\[ \vec{r} = \hat{i} + 2\hat{j} - \hat{k} + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \]

is parallel to the plane

\[ \vec{r} \cdot (3\hat{i} - 2\hat{j} + m\hat{k}) = 5, \]

we will proceed with the following steps:

### Step 1: Identify the direction vector of the line
The equation of the line can be expressed in the form:

\[ \vec{r} = \vec{a} + \lambda \vec{b} \]

where \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = 2\hat{i} + \hat{j} + 2\hat{k} \).

### Step 2: Identify the normal vector of the plane
The equation of the plane can be rewritten in the form:

\[ \vec{r} \cdot \vec{n} = d \]

where \( \vec{n} = 3\hat{i} - 2\hat{j} + m\hat{k} \) is the normal vector of the plane.

### Step 3: Set the condition for parallelism
For the line to be parallel to the plane, the direction vector \( \vec{b} \) must be perpendicular to the normal vector \( \vec{n} \). This means that their dot product must equal zero:

\[ \vec{b} \cdot \vec{n} = 0 \]

### Step 4: Calculate the dot product
Substituting \( \vec{b} \) and \( \vec{n} \):

\[ (2\hat{i} + \hat{j} + 2\hat{k}) \cdot (3\hat{i} - 2\hat{j} + m\hat{k}) = 0 \]

Calculating the dot product:

\[ 2 \cdot 3 + 1 \cdot (-2) + 2 \cdot m = 0 \]

This simplifies to:

\[ 6 - 2 + 2m = 0 \]

### Step 5: Solve for \( m \)
Now, we can simplify the equation:

\[ 4 + 2m = 0 \]

Rearranging gives:

\[ 2m = -4 \]

Dividing by 2:

\[ m = -2 \]

### Conclusion
The value of \( m \) for which the line is parallel to the plane is

\[ \boxed{-2} \]