If A(0,4,0),B(0,0,3),and C(0,4,3) are the ver | vedclass

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(C) The vertices are $A(0,4,0)$,$B(0,0,3)$,and $C(0,4,3)$.Calculate the side lengths:$a = BC = \sqrt{(0-0)^2 + (4-0)^2 + (3-3)^2} = \sqrt{0+16+0} = 4$

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$(0,3,2)$

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(C) The vertices are $A(0,4,0)$,$B(0,0,3)$,and $C(0,4,3)$.Calculate the side lengths:$a = BC = \sqrt{(0-0)^2 + (4-0)^2 + (3-3)^2} = \sqrt{0+16+0} = 4$.$b = AC = \sqrt{(0-0)^2 + (4-4)^2 + (3-0)^2} = \sqrt{0+0+9} = 3$.$c = AB = \sqrt{(0-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{0+16+9} = 5$.The incentre $I(x, y, z)$ is given by the formula:$I = \left( \frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}, \frac{az_A + bz_B + cz_C}{a+b+c} \right)$.Substituting the values:$x = \frac{4(0) + 3(0) + 5(0)}{4+3+5} = 0$.$y = \frac{4(4) + 3(0) + 5(4)}{4+3+5} = \frac{16+0+20}{12} = \frac{36}{12} = 3$.$z = \frac{4(0) + 3(3) + 5(3)}{4+3+5} = \frac{0+9+15}{12} = \frac{24}{12} = 2$.Thus,the incentre is $(0,3,2)$.

If $A(0,4,0)$,$B(0,0,3)$,and $C(0,4,3)$ are the vertices of $\Delta ABC$,then its incentre is

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