The plane 2x + 3y + 4z = 1 meets the X-axis a | vedclass

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(C) The equation of the plane is $2x + 3y + 4z = 1$. To find the intersection with the $X$-axis,set $y = 0$ and $z = 0$: $2x = 1 \implies x = \frac{1}

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$(\frac{1}{6}, \frac{1}{9}, \frac{1}{12})$

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(C) The equation of the plane is $2x + 3y + 4z = 1$. To find the intersection with the $X$-axis,set $y = 0$ and $z = 0$: $2x = 1 \implies x = \frac{1}{2}$. Thus,$A = (\frac{1}{2}, 0, 0)$. To find the intersection with the $Y$-axis,set $x = 0$ and $z = 0$: $3y = 1 \implies y = \frac{1}{3}$. Thus,$B = (0, \frac{1}{3}, 0)$. To find the intersection with the $Z$-axis,set $x = 0$ and $y = 0$: $4z = 1 \implies z = \frac{1}{4}$. Thus,$C = (0, 0, \frac{1}{4})$. The centroid of $	riangle 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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$. Substituting the coordinates: Centroid $= (\frac{1/2 + 0 + 0}{3}, \frac{0 + 1/3 + 0}{3}, \frac{0 + 0 + 1/4}{3}) = (\frac{1}{6}, \frac{1}{9}, \frac{1}{12})$.

The plane $2x + 3y + 4z = 1$ meets the $X$-axis at $A$,the $Y$-axis at $B$,and the $Z$-axis at $C$. Then the centroid of $\triangle ABC$ is:

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