Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The equation of the line passing through $(1,1,1)$ and $(2,2,2)$ is given by

$\frac{x-1}{2-1}=\frac{y-1}{2-1}=\frac{z-1}{2-1}$

$\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}=k$

Any point on this line is of the form $(k+1,k+1,k+1)$.

Since this point lies on the plane $x+y+z=9$, we substitute the coordinates into the equation of the plane:

$(k+1)+(k+1)+(k+1)=9$

$3k+3=9$

$3k=6$

$k=2$

Substituting $k=2$, the coordinates of the point of intersection are $(2+1,2+1,2+1)=(3,3,3)$.

Answer: $(3,3,3)$