Let three vectors ,$$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:

<p>To solve this, let's start with the given vector equation:</p> <p>$\vec{c} = \vec{a} - \vec{b}$</p> <p>Given vectors are:</p> <p>$\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$</p> <p>$\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$</p> <p>Then, the vector $\overrightarrow{\mathrm{c}}$ is:</p> <p>$$\overrightarrow{\mathrm{c}} = (\alpha \hat{i} + 4 \hat{j} + 2 \hat{k}) - (5 \hat{i} + 3 \hat{j} + 4 \hat{k})$$</p> <p>$$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k}$$</p> <p>$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$</p> <p>The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of the cross product of $\vec{a}$ and $\vec{b}$, divided by 2:</p> <p>$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 5 \sqrt{6}$</p> <p>This implies:</p> <p>$|\vec{a} \times \vec{b}| = 10 \sqrt{6}$</p> <p>Let's find $\vec{a} \times \vec{b}$:</p> <p>$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = \hat{i}(4 \cdot 4 - 2 \cdot 3) - \hat{j}(\alpha \cdot 4 - 2 \cdot 5) + \hat{k}(\alpha \cdot 3 - 4 \cdot 5)$$</p> <p>$$\vec{a} \times \vec{b} = \hat{i}(16 - 6) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$$</p> <p>$$\vec{a} \times \vec{b} = \hat{i}(10) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$$</p> <p>Magnitude of $\vec{a} \times \vec{b}$:</p> <p>$|\vec{a} \times \vec{b}| = \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2}$</p> <p>We know:</p> <p>$\sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} = 10 \sqrt{6}$</p> <p>Squaring both sides:</p> <p>$100 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600$</p> <p>$100 + 16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 600$</p> <p>$25\alpha^2 - 200\alpha + 600 = 600$</p> <p>$25\alpha^2 - 200\alpha = 0$</p> <p>$\alpha^2 - 8\alpha = 0$</p> <p>$\alpha(\alpha - 8) = 0$</p> <p>Since $\alpha$ is a positive real number:</p> <p>$\alpha = 8$</p> <p>Then, the vector $\vec{c}$ is:</p> <p>$\vec{c} = (8 - 5)\hat{i} + 1\hat{j} - 2\hat{k}$</p> <p>$\vec{c} = 3\hat{i} + 1\hat{j} - 2\hat{k}$</p> <p>Magnitude squared of $\vec{c}$ is:</p> <p>$|\vec{c}|^2 = 3^2 + 1^2 + (-2)^2$</p> <p>$|\vec{c}|^2 = 9 + 1 + 4$</p> <p>$|\vec{c}|^2 = 14$</p> <p>Therefore, the correct option is:</p> <p>Option A: 14</p>