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(B) Given $\vec{c} 	imes (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k}) 	imes \vec{d}$, which implies $\vec{c} 	imes (2\hat{i}+3\hat{j

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$1$

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(B) Given $\vec{c} 	imes (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k}) 	imes \vec{d}$, which implies $\vec{c} 	imes (2\hat{i}+3\hat{j}+4\hat{k}) + \vec{d} 	imes (2\hat{i}+3\hat{j}+4\hat{k}) = 0$. Thus, $(\vec{c}+\vec{d}) 	imes (2\hat{i}+3\hat{j}+4\hat{k}) = 0$. This means $\vec{c}+\vec{d}$ is parallel to $(2\hat{i}+3\hat{j}+4\hat{k})$. Let $\vec{c}+\vec{d} = \lambda(2\hat{i}+3\hat{j}+4\hat{k})$. Given $|\vec{c}+\vec{d}| = \sqrt{29}$, we have $|\lambda| \sqrt{2^2+3^2+4^2} = \sqrt{29}$, so $|\lambda| \sqrt{29} = \sqrt{29}$, which gives $\lambda = \pm 1$. Now, $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k}) = \lambda(2\hat{i}+3\hat{j}+4\hat{k}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k}) = \lambda(-14+6+12) = 4\lambda$. For $\lambda = 1$, the value is $4$, and for $\lambda = -1$, the value is $-4$. Thus, $\lambda_1 = 4$ and $\lambda_2 = -4$. The equation becomes $K^2x^2 + (K^2-5K+4)xy + (3K-2)y^2 - 8x + 12y - 4 = 0$. For this to represent a circle, the coefficient of $xy$ must be $0$ and the coefficients of $x^2$ and $y^2$ must be equal. $K^2-5K+4 = 0 \Rightarrow (K-1)(K-4) = 0 \Rightarrow K=1$ or $K=4$. $K^2 = 3K-2 \Rightarrow K^2-3K+2 = 0 \Rightarrow (K-1)(K-2) = 0 \Rightarrow K=1$ or $K=2$. The common value is $K=1$.

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