JEE Mathematics 2023 Question with Solution

Given: The foot of the perpendicular from the origin to plane $P$ is $(2, a, 4)$, and the volume of tetrahedron $OABC$ is $144$.

Find: Which given point does not lie on plane $P$.

A normal vector to the plane is along the perpendicular from the origin, so the plane can be written as

$$(2\hat{i}+a\hat{j}+4\hat{k})\cdot[(x-2)\hat{i}+(y-a)\hat{j}+(z-4)\hat{k}] = 0$$

Hence,

$$2x+ay+4z = 20+a^2$$

The intercepts on the coordinate axes are therefore

$$A\equiv\left(\frac{20+a^2}{2},0,0\right),\quad B\equiv\left(0,\frac{20+a^2}{a},0\right),\quad C\equiv\left(0,0,\frac{20+a^2}{4}\right)$$

Now using the volume of tetrahedron $OABC$,

$$\text{Volume} = \frac{1}{6}(OA)(OB)(OC) = 144$$

So,

$$\frac{1}{6}\left(\frac{20+a^2}{2}\right)\left(\frac{20+a^2}{a}\right)\left(\frac{20+a^2}{4}\right)=144$$

which gives

$$(20+a^2)^3 = 144\times 48\times a$$

From the working, we get

$$a=2$$

Therefore, the plane is

$$2x+2y+4z=24$$

or

$$x+y+2z=12$$

Now check the options. For the point $(3,0,4)$,

$$3+0+2(4)=11\neq 12$$

So this point is not on the plane.

Therefore, the point not lying on $P$ is $(3,0,4)$. The solution states the correct option as B, which disagrees with the listed options; among the given options this point corresponds to option C.

Use the fact that the perpendicular from the origin gives the normal direction of the plane. After obtaining

$$x+y+2z=12$$

substitute each option:

• $(2,2,4)$ gives $2+2+8=12$, so it lies on the plane.
• $(0,4,4)$ gives $0+4+8=12$, so it lies on the plane.
• $(3,0,4)$ gives $3+0+8=11$, so it does not lie on the plane.
• $(0,6,6)$ gives $0+6+12=18$, so it also does not lie on the plane.

Thus the provided solution is internally inconsistent with the options, because both $(3,0,4)$ and $(0,6,6)$ fail the plane equation. Since the solution explicitly concludes $(3,0,4)$, the most defensible mapped answer is B as declared in the solution, even though that point is listed as option C.