If the plane 2x + 3y + 5z = 1 intersects the | vedclass

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(C) The given equation of the plane is $2x + 3y + 5z = 1$. To find the intercepts on the coordinate axes,we rewrite the equation in the intercept form

Vedclass · Accepted Answer

$\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)$

Vedclass · Answer

(C) The given equation of the plane is $2x + 3y + 5z = 1$. To find the intercepts on the coordinate axes,we rewrite the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. This gives $\frac{x}{(1/2)} + \frac{y}{(1/3)} + \frac{z}{(1/5)} = 1$. Thus,the intercepts on the $X, Y, Z$ axes are $a = 1/2$,$b = 1/3$,and $c = 1/5$ respectively. The coordinates of the points $A, B, C$ are $A = (1/2, 0, 0)$,$B = (0, 1/3, 0)$,and $C = (0, 0, 1/5)$. The centroid of $	riangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$. Substituting the coordinates,we get $\left(\frac{1/2 + 0 + 0}{3}, \frac{0 + 1/3 + 0}{3}, \frac{0 + 0 + 1/5}{3}ight) = \left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}ight)$. Therefore,the correct option is $C$.

If the plane $2x + 3y + 5z = 1$ intersects the coordinate axes at the points $A, B, C$,then the centroid of $\triangle ABC$ is

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