Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane can be written in intercept form as $\frac{x}{2}+\frac{y}{-3}+\frac{z}{-5}=1$.

The intercepts on the coordinate axes are $a=2$, $b=-3$, and $c=-5$.

The points where the plane cuts the axes are $A(2,0,0)$, $B(0,-3,0)$, and $C(0,0,-5)$.

The area of the triangle formed by the intercepts on the coordinate axes is given by $\frac{1}{2}\sqrt{(ab{)}^2+(bc{)}^2+(ca{)}^2}$.

Substituting the values of $a$, $b$, and $c$:

$ab=2\times (-3)=-6$

$bc=(-3)\times (-5)=15$

$ca=(-5)\times 2=-10$

Area $=\frac{1}{2}\sqrt{(-6{)}^2+(15{)}^2+(-10{)}^2}$

Area $=\frac{1}{2}\sqrt{36+225+100}$

Area $=\frac{1}{2}\sqrt{361}$

Area $=\frac{19}{2}$ sq. units.

Answer: $\frac{19}{2}$ sq. units