The equation of the plane which makes with coordinate axes, a triangle witih centroid `(alpha,beta,gamma)` is given by

Options

A. `alphax+betay+gammaz=3`
B. `alphax+betay+gammaz=1`
C. `(x)/(alpha)+(y)/(beta)+(z)/(gamma)=3`
D. `(x)/(alpha)+(y)/(beta)+(z)/(gamma)=1`

Answer

C

Explanation

Let the vertices of triangle are A(a,0,0), B (0,b,0) and C(0,0,c) and the equation of plane is
`(x)/(a)+(y)/(b)+(z)/(c)=1` . . .(i)
`because` Centroid of `DeltaABC` is `(alpha,beta,gamma)`.
`:." "(a+0+0)/(3)=alpha`
`rArr" "a=3alpha` . . . (ii)
`and" "(0+b+0)/(3)=betarArrb=3beta` . . .(iii)
Similarly, `(0+0+c)/(3)=gamma`
`rArr" "c=3lamda` . . . (iv)
Using Eqs. (ii), (iii) and (iv) in Eq. (i), we get
`(x)/(3alpha)+(y)/(3beta)+(z)/(3gamma)=1`
`rArr" "(x)/(alpha)+(y)/(beta)+(z)/(gamma)=3`
which is the equation of required plane.