The system of linear equations $x+y+z=6$ $2 \mathrm{x}+5 \mathrm{y}+\mathrm{az}=36$ $x+2 y+3 z=b$ has

(B) infinitely many solutions for $a=8$ and $b=14$

(A) unique solution for $\mathrm{a}=8$ and $\mathrm{b}=16$

(C) infinitely many solutions for $a=8$ and $b=16$

(D) unique solution for $\mathrm{a}=8$ and $\mathrm{b}=14$

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Matrices

The system of linear equations

x + y + z = 6

2 x + 5 y + az = 36

x + 2 y + 3 z = b

has

Answer: (B)

The coefficient determinant is

D = 121152 1 a 3 = 8 - a

, which vanishes only at

a = 8

, so no unique solution exists there. With

a = 8

, the third equation's left side equals

\frac{1}{3}

(first)

+ \frac{1}{3}

(second), so consistency requires

b = \frac{1}{3} (6) + \frac{1}{3} (36) = 14

. Hence for

a = 8

and

b = 14

the planes meet in a line and the system has infinitely many solutions.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option