Find the angle between the line `vecr = (2hati+hatj-hatk)+lambda(2hati+2hatj+hatk)` and the plane `vecr.(6hati-3hatj+2hatk)+1=0`.

Explanation

To find the angle between the line given by the vector equation \(\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} + 2\hat{j} + \hat{k})\) and the plane defined by the equation \(\vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) + 1 = 0\), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line can be expressed in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
where \(\vec{a} = 2\hat{i} + \hat{j} - \hat{k}\) is a point on the line, and \(\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}\) is the direction vector of the line.
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