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(B) Given $\overrightarrow{c} = \overrightarrow{a} - \overrightarrow{b}$.Substituting the vectors,we get $\overrightarrow{c} = (\alpha - 5)\hat{i} + (

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$14$

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(B) Given $\overrightarrow{c} = \overrightarrow{a} - \overrightarrow{b}$.Substituting the vectors,we get $\overrightarrow{c} = (\alpha - 5)\hat{i} + (4 - 3)\hat{j} + (2 - 4)\hat{k} = (\alpha - 5)\hat{i} + 1\hat{j} - 2\hat{k}$.So,$x = \alpha - 5$,$y = 1$,and $z = -2$.The area of the triangle formed by vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\frac{1}{2} |\overrightarrow{a} 	imes \overrightarrow{b}|$.Since $\overrightarrow{c} = \overrightarrow{a} - \overrightarrow{b}$,the area is also $\frac{1}{2} |\overrightarrow{a} 	imes \overrightarrow{c}| = 5\sqrt{6}$.$\overrightarrow{a} 	imes \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 4 & 2 \ \alpha-5 & 1 & -2 \end{vmatrix} = \hat{i}(-8 - 2) - \hat{j}(-2\alpha - 2(\alpha - 5)) + \hat{k}(\alpha - 4(\alpha - 5))$$= -10\hat{i} - \hat{j}(-2\alpha - 2\alpha + 10) + \hat{k}(\alpha - 4\alpha + 20) = -10\hat{i} - (10 - 4\alpha)\hat{j} + (20 - 3\alpha)\hat{k}$.$|\overrightarrow{a} 	imes \overrightarrow{c}|^2 = (-10)^2 + (4\alpha - 10)^2 + (20 - 3\alpha)^2 = (10\sqrt{6} 	imes 2)^2 = 400 	imes 6 = 2400$.$100 + 16\alpha^2 - 80\alpha + 100 + 400 - 120\alpha + 9\alpha^2 = 2400$.$25\alpha^2 - 200\alpha + 600 = 2400 \Rightarrow 25\alpha^2 - 200\alpha - 1800 = 0$.Dividing by $25$,$\alpha^2 - 8\alpha - 72 = 0$. Wait,let's re-evaluate the cross product magnitude.$|\overrightarrow{a} 	imes \overrightarrow{c}| = 10\sqrt{6} \Rightarrow |\overrightarrow{a} 	imes \overrightarrow{c}|^2 = 600$.$100 + (4\alpha - 10)^2 + (20 - 3\alpha)^2 = 600$.$100 + 16\alpha^2 - 80\alpha + 100 + 400 - 120\alpha + 9\alpha^2 = 600$.$25\alpha^2 - 200\alpha + 600 = 600 \Rightarrow 25\alpha^2 - 200\alpha = 0$.$25\alpha(\alpha - 8) = 0$. Since $\alpha > 0$,$\alpha = 8$.Then $x = 8 - 5 = 3, y = 1, z = -2$.$|\overrightarrow{c}|^2 = x^2 + y^2 + z^2 = 3^2 + 1^2 + (-2)^2 = 9 + 1 + 4 = 14$.

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