Let a, b, c be in A.P. and the line $ax + by + c = 0$ passes through a fixed point P. For the system of equations $$\begin{aligned} x + y + z &= 6 \\ 2x + 5y + \alpha z &= \beta \\ x + 2y + 3z &= 4 \end{aligned}$$ to have infinite solutions, let the point be Q$(\alpha, \beta)$. Then $PQ^2$ is equal to ______.

Matrices — JEE Main · 2024 · 29 Jan 2024 · Shift 2 Question with Solution

Let a, b, c be in A.P. and the line

a x + b y + c = 0

passes through a fixed point P. For the system of equations

x + y + z 2 x + 5 y + α z x + 2 y + 3 z = 6 = β = 4

to have infinite solutions, let the point be Q

(α, β)

. Then

P Q^{2}

is equal to ______.

Answer: 113

a, b, c

in A.P.

\Rightarrow 2 b = a + c \Rightarrow a - 2 b + c = 0

, which says the line

a x + b y + c = 0

always satisfies

a (1) + b (- 2) + c (1) = 0

. So it always passes through the fixed point

P = (1, - 2)

.

For the system

x + y + z = 6, 2 x + 5 y + α z = β, x + 2 y + 3 z = 4

to have infinitely many solutions, the coefficient determinant must vanish:

121152 1 α 3 = 1 (15 - 2 α) - 1 (6 - α) + 1 (4 - 5) = 15 - 2 α - 6 + α - 1 = 8 - α = 0 \Rightarrow α = 8

.

For consistency (infinite, not zero, solutions), replacing the

x

-column with the RHS must also give a vanishing determinant:

6 β 4 152183 = 6 (15 - 16) - 1 (3 β - 32) + 1 (2 β - 20) = - 6 - 3 β + 32 + 2 β - 20 = 6 - β = 0 \Rightarrow β = 6

.

So

Q = (α, β) = (8, 6)

P Q^{2} = (8 - 1)^{2} + (6 - (- 2))^{2} = 49 + 64 = 113

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer