Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane is $2x-3y-6z=49$.

The direction ratios of the normal to the plane are $(2,-3,-6)$.

The equation of the line passing through the origin and perpendicular to the plane is $\frac{x}{2}=\frac{y}{-3}=\frac{z}{-6}=k$.

Any point on this line is of the form $(2k,-3k,-6k)$.

Since this point is the foot of the perpendicular, it must lie on the plane. Substituting these coordinates into the equation of the plane:

$2(2k)-3(-3k)-6(-6k)=49$

$4k+9k+36k=49$

$49k=49\Rightarrow k=1$

Therefore, the coordinates of the foot of the perpendicular are $(2(1),-3(1),-6(1))=(2,-3,-6)$.

Answer: $(2,-3,-6)$