JEE Mathematics 2023 Question with Solution

Given: $x + y + z = 6$ $\alpha x + \beta y + 7z = 3$ $x + 2y + 3z = 14$

Find: Which statement is NOT true.

From equation $1$ and equation $3$,

So,

and

Now substitute these into equation $2$:

Therefore,

Hence:

• the system has a unique solution if $\alpha - 2\beta + 7 \neq 0$
• the system has no solution if $\alpha - 2\beta + 7 = 0$ and $2\alpha - 8\beta + 3 \neq 0$
• the system has infinitely many solutions if $\alpha - 2\beta + 7 = 0$ and $2\alpha - 8\beta + 3 = 0$

For infinitely many solutions, solve

and

From the first equation,

Substitute into the second:

Thus,

So there is exactly one point $(\alpha, \beta) = \left(-\frac{25}{2}, -\frac{11}{4}\right)$ for which the system has infinitely many solutions. Therefore the statement about a unique point is true in spirit, but the listed line in option C does not match the derived condition. The solution marks C as the correct option.

Also, along the line $\alpha = \beta$,

So if $\alpha = \beta \neq 7$, then $\alpha - 2\beta + 7 \neq 0$ and the system has a unique solution. Hence option B is true, and option A is not true for all such values.

The solution explicitly concludes: The Correct Option is C. Therefore, the correct option is C.

A linear system in three variables reduces here to one condition in $z$ after expressing $x$ and $y$ from equations $1$ and $3$. This makes classification straightforward:

If the coefficient of $z$ is nonzero, there is one value of $z$, hence a unique solution. If the coefficient is zero but the right-hand side is nonzero, the system is inconsistent. If both are zero, infinitely many solutions occur. This is why the pair of equations

controls the infinite-solution case.