The foot of the perpendicular drawn from the origin to the plane $x+y+z=3$ is

Given equation of plane $x+y+z=3$ Let point $P$ is $(0,0,0)$ and point $Q$ is the foot of perpendicular drawn from point $P$ on the plane. Since, PQ is perpendicular to the plane, so direction ratio of line $\mathrm{PQ}\langle 1,1,1\rangle$. $\therefore \quad$ Equation of line $P Q$, $\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}=\lambda$ $\therefore(x, y, z)=(\lambda, \lambda, \lambda)$ As point Q lies on the given plane. $\therefore \lambda+\lambda+\lambda=3 \Rightarrow \lambda=1$ Hence, Co-ordinate of point $\mathrm{Q}=(1,1,1)$.