Let $\lambda \in \mathbb{Z}, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. Let $\vec{c}$ be a vector such that $$(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$$ and $\vec{b} \cdot \vec{c}=-20$. Then $|\vec{c} \times(\lambda \hat{i}+\hat{j}+\hat{k})|^{2}$ is equal to :

The given vectors are : <br/><br/>$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ <br/><br/>$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$ <br/><br/>We are given that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$ which implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. So, $\vec{c}$ is in the direction of $\vec{a} + \vec{b}$. <br/><br/>Let's denote $\vec{c} = \alpha (\vec{a} + \vec{b})$. <br/><br/>Substituting values for $\vec{a}$ and $\vec{b}$, we get : <br/><br/>$\vec{c} = \alpha((\lambda + 3) \hat{i} + \hat{k})$ <br/><br/>From the conditions $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$, we can form two equations : <br/><br/>$$\alpha \lambda(\lambda + 3) - \alpha = -17 \quad \text{and} \quad \alpha(3\lambda + 9 + 2) = -20.$$ <br/><br/>By solving the above equations, we find $\alpha = -1$ and $\lambda = 3$. <br/><br/>Substituting these values back into $\vec{c}$ gives $\vec{c} = -6 \hat{i} - \hat{k}$. <br/><br/>Next, we need to find $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$. <br/><br/>This simplifies to : <br/><br/>$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2$ <br/><br/>Calculate the cross product $\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})$ which gives : <br/><br/>$$ =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{array}\right|=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} $$ <br/><br/>Finally, the square of the magnitude of this vector is: <br/><br/>$$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2 = (1)^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46.$$