Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} imes(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) imes \overrightarrow{\mathrm{d}}$. If $\lambda_1, \lambda_2\left(\lambda_1\lambda_2 ight)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^2 x^2+\left(\mathrm{K}^2-5 \mathrm{~K}+\lambda_1 ight) x y+\left(3 \mathrm{~K}+\frac{\lambda_2}{2} ight) y^2-8 x+12 y+\lambda_2=0$ represents a circle, for K equal to :

Question

Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) 	imes \overrightarrow{\mathrm{d}}$. If $\lambda_1, \lambda_2\left(\lambda_1>\lambda_2ight)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^2 x^2+\left(\mathrm{K}^2-5 \mathrm{~K}+\lambda_1ight) x y+\left(3 \mathrm{~K}+\frac{\lambda_2}{2}ight) y^2-8 x+12 y+\lambda_2=0$ represents a circle, for K equal to :

Accepted Answer

D. 1. Rearrange the given cross-product condition to show that $\vec{c}+\vec{d}$ is parallel to the vector $2\hat{i}+3\hat{j}+4\hat{k}$.
Given
$$\vec{c}	imes(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})	imes\vec{d}$$
So
$$(\vec{c}+\vec{d})	imes(2\hat{i}+3\hat{j}+4\hat{k})=\vec{0} \Rightarrow \vec{c}+\vec{d}=\lambda(2\hat{i}+3\hat{j}+4\hat{k})$$

2. Use the given magnitude $|\vec{c}+\vec{d}|=\sqrt{29}$ to find $\lambda$, noting that $|2\hat{i}+3\hat{j}+4\hat{k}|=\sqrt{29}$ as well.
$$\sqrt{29}=|\lambda|\sqrt{29} \Rightarrow \lambda=\pm1$$
So
$$\vec{c}+\vec{d}=\pm(2\hat{i}+3\hat{j}+4\hat{k})$$

3. Compute $(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$ for both signs to find $\lambda_1$ and $\lambda_2$.
$$(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})=\pm(-14+6+12)=\pm4$$
So
$$\lambda_1=4,\quad \lambda_2=-4$$

4. Apply the condition for the given second-degree equation to represent a circle: the coefficients of $x^2$ and $y^2$ must be equal, and the coefficient of $xy$ must be zero.
Given the equation
$$K^2x^2+(K^2-5K+\lambda_1)xy+\left(3K+\frac{\lambda_2}{2}ight)y^2-8x+12y+\lambda_2=0$$
So
$$K^2=3K+\frac{\lambda_2}{2}=3K-2 \quad	ext{and}\quad K^2-5K+\lambda_1=0 \Rightarrow K^2-5K+4=0$$

5. Solve each quadratic in $K$.
$$K^2-3K+2=0 \Rightarrow K=1,2$$
$$K^2-5K+4=0 \Rightarrow K=1,4$$

6. Take the common value of $K$ satisfying both equations.
$$K=1$$

Hence, the answer is **Option D (K = 1)**.