JEE Mathematics 2023 Question with Solution

Given: The plane passes through the points $A(-1, k, 0), B(2, k, -1), C(1, 1, 2)$ and is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$.

Find: The value of $\frac{k^2+1}{(k-1)(k-2)}$.

From the solution, the direction ratios used for the line are taken as $(1,1,-1)$ after rewriting it as

although this differs from the question statement, where the middle term is $\frac{2y+1}{2}$.

Using the points, the vectors are taken as

and

Their cross product gives a normal vector to the plane:

The solution then uses the condition involving the given line and concludes that the required expression simplifies to $\frac{13}{6}$. It explicitly states The Correct Option is A. Since the computed value $\frac{13}{6}$ corresponds to option D in the listed options, there is a discrepancy between the option label stated in the solution and the numerical value obtained there.

Therefore, based on the solution's stated option label, the answer is taken as A.

The solution is internally inconsistent:

1. It says the plane is parallel to the line in the question, but later says the line is perpendicular to the plane.
2. It rewrites $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$ as $\frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{-1}$.
3. It concludes the value is $\frac{13}{6}$, which matches option D, but the solution states The Correct Option is A.

By the instruction that the solution is the primary source for the answer, the option label from the solution is used.