JEE Main 2026 — 21 January, Morning Shift — Question 7

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 $(\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}) \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$, 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 :