If the plane frac{x}{2} + frac{y}{3} + frac{z | vedclass

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(D) The given plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$. The intercepts on the axes are $a=2, b=3, c=3$. Thus,the coordinates of the vert

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None of these

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(D) The given plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$. The intercepts on the axes are $a=2, b=3, c=3$. Thus,the coordinates of the vertices are $A(2, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 3)$. The vectors forming the sides are $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 3)$. The area of $\Delta ABC$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. Calculating the cross product: $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 3 \end{vmatrix} = \hat{i}(9 - 0) - \hat{j}(-6 - 0) + \hat{k}(0 - (-6)) = 9\hat{i} + 6\hat{j} + 6\hat{k}$. The magnitude is $|\vec{AB} 	imes \vec{AC}| = \sqrt{9^2 + 6^2 + 6^2} = \sqrt{81 + 36 + 36} = \sqrt{153} = 3\sqrt{17}$. Therefore,the area is $\frac{1}{2} 	imes 3\sqrt{17} = \frac{3\sqrt{17}}{2}$ sq. units. Since this value is not among the options,the correct choice is $D$.

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

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