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

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(B) The equation of the plane is $\frac{x}{2} - \frac{y}{3} - \frac{z}{5} = 1$. Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \f

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$\frac{\sqrt{1529}}{6}$ sq. units

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(B) The equation of the plane is $\frac{x}{2} - \frac{y}{3} - \frac{z}{5} = 1$. Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get the intercepts on the axes as $a = 2, b = -3, c = -5$. Thus,the coordinates of the points are $A(2, 0, 0)$,$B(0, -3, 0)$,and $C(0, 0, -5)$. The area of triangle $ABC$ with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$. Substituting the values: $	ext{Area} = \frac{1}{2} \sqrt{(2 	imes -3)^2 + (-3 	imes -5)^2 + (-5 	imes 2)^2}$. $	ext{Area} = \frac{1}{2} \sqrt{(-6)^2 + (15)^2 + (-10)^2} = \frac{1}{2} \sqrt{36 + 225 + 100} = \frac{1}{2} \sqrt{361} = \frac{19}{2}$ sq. units.

If the plane $\frac{x}{2} - \frac{y}{3} - \frac{z}{5} = 1$ cuts the coordinate axes at points $A, B,$ and $C$ respectively,then the area of the triangle $ABC$ is:

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