JEE Mathematics 2025 Question with Solution

Given:

$$x + y + 2z = 6$$ $$2x + 3y + az = a + 1$$ $$-x - 3y + bz = 2b$$

Find: The value of $7a + 3b$ when the system has infinitely many solutions.

For infinitely many solutions, the system must satisfy the consistency condition

$$\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$$

and in particular the determinant of the coefficient matrix must be zero.

The coefficient determinant is

$$\begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0$$

Expanding,

$$(3b + 3a) - (2b + a) + 2(-6 + 3) = 0$$ $$b + 2a - 6 = 0$$

So,

$$2a + b = 6$$

Now compare two derived equations from the original system. Subtract the first equation from the second:

$$(2x + 3y + az) - (x + y + 2z) = (a + 1) - 6$$ $$x + 2y + (a - 2)z = a - 5$$

Subtract the first equation from the third:

$$(-x - 3y + bz) - (x + y + 2z) = 2b - 6$$ $$-2x - 4y + (b - 2)z = 2b - 6$$

Dividing by $-2$,

$$x + 2y - \frac{b - 2}{2}z = 3 - b$$

For infinitely many solutions, these two equations must represent the same relation. Hence,

$$a - 5 = 3 - b$$

which gives

$$a + b = 8$$

Using

$$2a + b = 6$$

and

$$a + b = 8$$

subtract the second equation from the first:

$$a = -2$$

Then

$$b = 10$$

Therefore,

$$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16$$

So the correct option is B.