JEE Mathematics 2023 Question with Solution

Given: One vertex is $A(0,2,\alpha)$. The other two vertices $B$ and $C$ lie on the line $x + \alpha/5 = (y-1)/2 = (z+4)/3$. The area of $\triangle ABC$ is $21$ square units and $BC = 2\sqrt{21}$.

Find: $\alpha^2$.

Take the line in symmetric form as

So a general point on the line is

The perpendicular distance from $A$ to this line is the altitude of the triangle on base $BC$.

Using the area formula,

we get

Hence

Choose a point on the line by taking $t=0$:

A direction vector of the line is

Then

Distance of point $A$ from the line through $B$ in direction $\vec d$ is

Now

which gives

Therefore,

Since

and $h=\sqrt{21}$, we have

So

Expanding,

Factoring,

Since $\alpha \in Z$, we get

Hence

Therefore, the correct option is A.

The solution concludes $\alpha^2=9$ and its header also lists Correct Answer: 9. This disagrees with the answer key and option list entry (3) 25. the answer is taken as A.

Because $B$ and $C$ lie on the same line, the base $BC$ is along direction $(5,2,3)$. The area condition together with the given base length directly gives the altitude:

Now compute the perpendicular distance from $A(0,2,\alpha)$ to the line $\frac{x+\alpha}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ using one point $(-\alpha,1,-4)$ on the line and direction vector $(5,2,3)$. Equating that distance to $\sqrt{21}$ yields the quadratic in $\alpha$, whose only integer solution is $\alpha=3$. Hence $\alpha^2=9$.