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The incenter and centroid of the triangle, whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$, and $C \equiv(0,3,4)$, are respectively given by
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The correct answer is:
$(0,2,3),\left(0,2, \frac{8}{3}\right)$
$A \equiv(0,3,0), B \equiv(0,0,4)$, and $C \equiv(0,3,4)$
$a=B C=3$
$b=C A=4$
$c=A B=5$.

$\begin{aligned} & \text { Now, Incentre } \equiv\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}, \frac{a z_1+b z_2+c z_3}{a+b+c}\right) \\ & \equiv\left(\frac{3 \times 0+4 \times 0+5 \times 0}{3+4+5}, \frac{3 \times 3+4 \times 0+5 \times 3}{3+4+5}, \frac{3 \times 0+4 \times 4+5 \times 4}{3+4+5}\right) \\ & \equiv(0,2,3) \\ & \text { and centroid } \equiv\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\end{aligned}$
$\begin{aligned} & \equiv\left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) \\ & \equiv\left(0,2, \frac{8}{3}\right)\end{aligned}$
$a=B C=3$
$b=C A=4$
$c=A B=5$.

$\begin{aligned} & \text { Now, Incentre } \equiv\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}, \frac{a z_1+b z_2+c z_3}{a+b+c}\right) \\ & \equiv\left(\frac{3 \times 0+4 \times 0+5 \times 0}{3+4+5}, \frac{3 \times 3+4 \times 0+5 \times 3}{3+4+5}, \frac{3 \times 0+4 \times 4+5 \times 4}{3+4+5}\right) \\ & \equiv(0,2,3) \\ & \text { and centroid } \equiv\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\end{aligned}$
$\begin{aligned} & \equiv\left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) \\ & \equiv\left(0,2, \frac{8}{3}\right)\end{aligned}$
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