JEE Mathematics 2024 Question with Solution

Given: $\vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}$, $\vec{b} = -\hat{i} + \hat{k}$, $\vec{c} = \beta \hat{j} - \hat{k}$ and $\alpha\beta = -6$.

Find: The value asked in the question, using the condition that the area of the parallelogram having diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$.

First form the diagonals:

For a parallelogram, area in terms of diagonals is

Now compute the cross product:

Hence,

Given area is $\frac{\sqrt{21}}{2}$, so

Thus,

Using $\alpha\beta = -6$,

Now test integer pairs satisfying $\alpha\beta = -6$: $(2,-3), (-2,3), (3,-2), (-3,2), (6,-1), (-6,1), (1,-6), (-1,6)$. From the working, the valid pairs are $(\alpha_1, \beta_1) = (3,-2)$ and $(\alpha_2, \beta_2) = (-3,2)$. Then the final evaluation shown in the solution is

Therefore, the correct option is D, i.e. $19$.

Note: The source solution concludes with $19$ and uses the expression $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2$ in its final substitution, whereas the question text displays $\alpha_1 + \beta_1 - \alpha_2\beta_2$. The extracted answer follows the solution, which explicitly marks option D as correct.

Given: The same vectors and area condition.

Find: The correct option from the source solution set.

A second approach on the solution's writes a different form of $\vec{b}$ and proceeds with a similar cross-product method. It again concludes that the correct option is $19$.

Because both solution blocks on the solution's contain internal inconsistencies in the statement being evaluated, the reliable extracted conclusion is the one common to both: the marked answer is Option D.