JEE Mathematics 2023 Question with Solution

Given: The point is $P(3,-2,-9)$ and the plane passes through $A(-1,-2,-3)$, $B(9,3,4)$ and $C(9,-2,1)$.

Find: The distance of the foot of the perpendicular $Q(\alpha,\beta,\gamma)$ from the origin.

First find the equation of the plane.

The normal vector is

So an equivalent normal vector is

Using point $A(-1,-2,-3)$, the equation of the plane is

which simplifies to

The perpendicular from $P$ to the plane is along the normal vector. Hence the line through $P(3,-2,-9)$ is

So,

Since $Q$ lies on the plane, substitute these values into the plane equation:

Therefore,

Now the distance of $Q$ from the origin is

Therefore, the correct option is D.

Given: The foot of the perpendicular from $P(3,-2,-9)$ to the plane through the three given points is $Q$.

Find: The distance $OQ$.

The key idea is that the foot of the perpendicular lies on the line through $P$ in the direction of the plane's normal vector. Once the plane normal $(2,3,-5)$ is obtained from $\overrightarrow{AB}\times\overrightarrow{AC}$, write

and impose the plane equation

to get $t=-1$ immediately.

Then

So,

Therefore, the correct option is D.