Vector Algebra Part 3

Concept:The cross-product condition shows $\vec{c}+\vec{d}$ is parallel to $2\hat{i}+3\hat{j}+4\hat{k}$, and a second-degree equation represents a circle when coefficients of $x^2$ and $y^2$ are equal and the $xy$ term is absent.Explanation:Given $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$.This simplifies to $(\vec{c}+\vec{d})\times(2\hat{i}+3\hat{j}+4\hat{k})=0$.Hence $\vec{c}+\vec{d}=\lambda(2\hat{i}+3\hat{j}+4\hat{k})$.Magnitude: $|\vec{c}+\vec{d}|=|\lambda|\sqrt{4+9+16}=|\lambda|\sqrt{29}$.Given $|\vec{c}+\vec{d}|=\sqrt{29}$, so $|\lambda|=1$, i.e., $\lambda=\pm1$.Thus $\vec{c}+\vec{d}=\pm(2\hat{i}+3\hat{j}+4\hat{k})$.Compute $(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})=\pm(-14+6+12)=\pm4$.So $\lambda_1=4$, $\lambda_2=-4$ (since $\lambda_1>\lambda_2$).Given equation: $K^2x^2+(K^2-5K+\lambda_1)xy+(3K+\frac{\lambda_2}{2})y^2-8x+12y+\lambda_2=0$.For a circle: coefficient of $x^2$ equals coefficient of $y^2$, and coefficient of $xy$ is zero.Therefore $K^2=3K+\frac{\lambda_2}{2}$ and $K^2-5K+\lambda_1=0$.Substitute $\lambda_2=-4$: $K^2=3K-2\Rightarrow K^2-3K+2=0\Rightarrow K=1,2$.Substitute $\lambda_1=4$: $K^2-5K+4=0\Rightarrow K=1,4$.Common value: $K=1$.Answer:$K=1$, which corresponds to option D.