If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}$, the magnitude of $[(\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})]$ is

Given $\mathrm{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\mathrm{b} = 3\hat{i} + 2\hat{j} - \hat{k}$, we determine $|(\mathrm{a} + 3\mathrm{b}) \cdot (2\mathrm{a} - \mathrm{b})|$.

The first vector sum is $\mathrm{a} + 3\mathrm{b} = (1\hat{i} + 1\hat{j} + 2\hat{k}) + (9\hat{i} + 6\hat{j} - 3\hat{k}) = 10\hat{i} + 7\hat{j} - \hat{k}$

The second vector is $2\mathrm{a} - \mathrm{b} = (2\hat{i} + 2\hat{j} + 4\hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k}) = -\hat{i} + 5\hat{k}$

Taking their dot product: $(10)(-1) + (7)(0) + (-1)(5) = -15$

The magnitude of this scalar is $|-15| = 15$

Final answer: $15$