The area of a triangle with vertices (1, 2, 0 | vedclass

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(B) Let the vertices of the triangle be $A(1, 2, 0)$,$B(1, 0, a)$,and $C(0, 3, 1)$.We know that the area of a triangle with vertices $A, B, C$ is give

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$2$,-$4$

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(B) Let the vertices of the triangle be $A(1, 2, 0)$,$B(1, 0, a)$,and $C(0, 3, 1)$.We know that the area of a triangle with vertices $A, B, C$ is given by $\frac{1}{2} |\vec{BA} 	imes \vec{BC}|$.First,we find the vectors $\vec{BA}$ and $\vec{BC}$:$\vec{BA} = (1-1)\hat{i} + (2-0)\hat{j} + (0-a)\hat{k} = 2\hat{j} - a\hat{k}$$\vec{BC} = (0-1)\hat{i} + (3-0)\hat{j} + (1-a)\hat{k} = -\hat{i} + 3\hat{j} + (1-a)\hat{k}$Now,calculate the cross product $\vec{BA} 	imes \vec{BC}$:$\vec{BA} 	imes \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & -a \ -1 & 3 & 1-a \end{vmatrix}$$= \hat{i}(2(1-a) - (-a)(3)) - \hat{j}(0(1-a) - (-a)(-1)) + \hat{k}(0(3) - 2(-1))$$= \hat{i}(2 - 2a + 3a) - \hat{j}(-a) + \hat{k}(2) = (a+2)\hat{i} + a\hat{j} + 2\hat{k}$The magnitude is $|\vec{BA} 	imes \vec{BC}| = \sqrt{(a+2)^2 + a^2 + 2^2} = \sqrt{a^2 + 4a + 4 + a^2 + 4} = \sqrt{2a^2 + 4a + 8}$.Given the area is $\sqrt{6}$,we have:$\frac{1}{2} \sqrt{2a^2 + 4a + 8} = \sqrt{6}$$\sqrt{2a^2 + 4a + 8} = 2\sqrt{6} = \sqrt{24}$$2a^2 + 4a + 8 = 24$$2a^2 + 4a - 16 = 0$$a^2 + 2a - 8 = 0$$(a+4)(a-2) = 0$Thus,$a = -4$ or $a = 2$.

The area of a triangle with vertices $(1, 2, 0)$,$(1, 0, a)$,and $(0, 3, 1)$ is $\sqrt{6}$ sq. units. Then the values of '$a$' are:

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