JEE Mathematics 2023 Question with Solution

Given: Point $A(4, 3, 1)$, plane $P : x - y + 2z + 3 = 0$, and the foot of the perpendicular from $A$ to the plane is $N$. Also, $B(5, \alpha, \beta)$ lies on the plane and the area of triangle $ABN$ is $3\sqrt{2}$.

Find: The value of $\alpha^2 + \beta^2 + \alpha \beta$.

Since $N$ is the foot of the perpendicular from $A$ to the plane $P$, the line $AN$ is along the normal to the plane. For the plane $x - y + 2z + 3 = 0$, a normal vector is

So the line through $A(4, 3, 1)$ and $N(x,y,z)$ is

Solving this with the plane equation gives

Hence,

Now,

Therefore,

Also,

So,

Using the area condition for triangle $ABN$,

From the given working, solving the resulting system gives

Now compute

Therefore, the required value is $7$.

Given: The extracted solution concludes that

Find: $\alpha^2 + \beta^2 + \alpha\beta$.

Substitute the values directly:

Thus, the numerical value answer is $7$.