Projection of \(\vec{b}\) on \(\vec{a}\) is \(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} = \vec{a}\). So \(\vec{b} \cdot \vec{a} = |\vec{a}|^2 = 1+1+2 = 4\). Thus \(b_1 + b_2 + 2 = 4\), giving \(b_1+b_2 = 2\). Condition \((\vec{a}+\vec{b}) \perp \vec{c}\): \((1+b_1)(5) + (1+b_2)(1) + (2\sqrt{2})(\sqrt{2}) = 0\), giving \(5+5b_1+1+b_2+4 = 0\), so \(10+5b_1+b_2 = 0\). From \(b_1+b_2=2\): \(b_2 = 2-b_1\), substituting: \(10+5b_1+2-b_1 = 0\), so \(4b_1 = -12\) and \(b_1 = -3\), giving \(b_2 = 5\). Thus \(\vec{b} = -3\hat{i}+5\hat{j}+\sqrt{2}\hat{k}\), and \(|\vec{b}| = \sqrt{9+25+2} = \sqrt{36} = 6\).