Question 2 — JEE Main 21 Jan 2026 (S I)

Question

Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \mathrm{k})=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \mathrm{k}) \times \overrightarrow{\mathrm{d}}$.
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 21 Jan 2026 S I]
(1) 4
(2) 1
(3) -1
(4) 2

Solution

$\overrightarrow{\mathrm{c}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=-\overrightarrow{\mathrm{d}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$
$\Rightarrow(\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=0$
$\therefore \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}} \| 2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}=\mathrm{p}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$
$|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=|\mathrm{p}| \sqrt{4+9+16}=\sqrt{29}$
$\Rightarrow|\mathrm{p}|=1 \Rightarrow \mathrm{p}= \pm 1$
$(\hat{\mathrm{c}}+\hat{\mathrm{d}}) \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\mathrm{p}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
$=\mathrm{p}[-14+6+12]=4 \mathrm{p}$
when $\mathrm{p}=1, \lambda_{1}=4$
when $p=-1, \lambda_{2}=-4$
Now equation of circle
$k^{2} x^{2}+\left(k^{2}-5 k+4\right) x y+(3 k-2) y^{2}-8 x+12 y-4=0$
coefficient of $\mathrm{xy}=0 \Rightarrow \mathrm{k}^{2}-5 \mathrm{k}+4=0$
$\Rightarrow(\mathrm{k}-1)(\mathrm{k}-4)=0 \Rightarrow \mathrm{k}=1,4$
coefficient of $\mathrm{x}^{2}=$ coefficient of $\mathrm{y}^{2}$
$\Rightarrow \mathrm{k}^{2}=3 \mathrm{k}-2 \Rightarrow \mathrm{k}^{2}-3 \mathrm{k}+2=0$
$\Rightarrow(\mathrm{k}-1)(\mathrm{k}-2)=0 \Rightarrow \mathrm{k}=1,2 \ldots(2)$
(1) $\cap(2)$
$\mathrm{k}=1$ only Answer(2)