Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^{2}=50$, then $|72-| \vec{b}+\left.\vec{c}\right|^{2} \mid$ is equal to __________.

<p>Given that $\vec{a} = \vec{b} \times \vec{c}$, we can find the magnitudes of $\vec{a}$ and $\vec{c}$:</p> <p>$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$ <br/><br/>$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$</p> <p>We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :</p> <p>$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$</p> <p>Plugging in the known values :</p> <p>$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$</p> <p>Solving for the sine of the angle between the vectors :</p> <p>$\sin\theta = \frac{1}{10}$</p> <p>Now we can find $|\vec{b} + \vec{c}|^2$ using the formula :</p> <p>$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$</p> <p>We have the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$, and we can use the relationship between sine and cosine: $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$.</p> <p>Substitute the values into the formula :</p> <p>$$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$$</p> <p>Finally, we need to find the absolute value of the difference :</p> <p>$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$</p>