Let \(\overrightarrow{\mathrm{c}}\) and \(\overrightarrow{\mathrm{d}}\) be vectors such that \(|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}\) and \(\overrightarrow{\mathrm{c}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{d}}\). If \(\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)\) are the possible values of \((\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})\), then the equation \(\mathrm{K}^{2} \mathrm{x}^{2}+\left(\mathrm{K}^{2}-5 \mathrm{~K}+\lambda_{1}\right) \mathrm{xy}+\left(3 \mathrm{~K}+\frac{\lambda_{2}}{2}\right) \mathrm{y}^{2}-8 \mathrm{x}+12 \mathrm{y}+\lambda_{2}=0\) represents a circle, for k equal to :

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Vector Algebra

Let

c

and

d

be vectors such that

∣ c + d ∣ = 29

and

c \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times d

. If

λ_{1}, λ_{2} (λ_{1} > λ_{2})

are the possible values of

(c + d) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})

, then the equation

K^{2} x^{2} + (K^{2} - 5 K + λ_{1}) xy + (3 K + \frac{λ _{2}}{2}) y^{2} - 8 x + 12 y + λ_{2} = 0

represents a circle, for k equal to :

Answer: (B)

With

v = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}

, the condition

c \times v = v \times d

rearranges to

v \times (c + d) = 0

, so

c + d = λ v

. Since

∣ c + d ∣ = 29 = ∣ v ∣

λ = \pm 1

. Then

(c + d) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = λ (- 14 + 6 + 12) = 4 λ

, giving

λ_{1} = 4, λ_{2} = - 4

. The equation

K^{2} x^{2} + (K^{2} - 5 K + 4) x y + (3 K - 2) y^{2} - 8 x + 12 y - 4 = 0

represents a circle only when the

x y

coefficient vanishes and the

x^{2}, y^{2}

coefficients are equal, i.e.

K^{2} - 5 K + 4 = 0

and

K^{2} = 3 K - 2

; their common root is

K = 1

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option