Find the angle between the planes `2x+y-2x+3=0 and vecr*(6hati+3hatj+2hatk)=5`.

Explanation

To find the angle between the two planes given by the equations \(2x + y - 2z + 3 = 0\) and \(\vec{r} \cdot (6\hat{i} + 3\hat{j} + 2\hat{k}) = 5\), we will follow these steps:

### Step 1: Identify the normal vectors of the planes

The normal vector of the first plane \(2x + y - 2z + 3 = 0\) can be derived from the coefficients of \(x\), \(y\), and \(z\). Thus, the normal vector \(N_1\) is:
\[ N_1 = \langle 2, 1, -2 \rangle \]
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