Solution
\[
\begin{aligned}
& \vec{c} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=-\vec{d} \times(2 \hat{\imath}+3 \hat{j}+4 \hat{k}) \\
& \Rightarrow(\vec{c}+\vec{d}) \times(2 \hat{\imath}+3 \hat{j}+4 \hat{k})=0 \\
& \therefore \vec{c}+\vec{d} \| 2 \hat{\imath}+3 \hat{j}+4 \hat{k} \\
& \vec{c}+\vec{d}=p(2 \hat{\imath}+3 \hat{j}+4 \hat{k}) \\
& |\vec{c}+\vec{d}|=|p| \sqrt{4+9+16}=\sqrt{29} \\
& \Rightarrow|p|=1 \Rightarrow p= \pm 1 \\
& (\hat{c}+\hat{d}) \cdot(-7 \hat{\imath}+2 \hat{j}+3 \hat{k})=p(2 \hat{\imath}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{\imath}+2 \hat{j}+3 \hat{k}) \\
& =p[-14+6+12]=4 p
\end{aligned}
\]
when $\mathrm{p}=1, \lambda_{1}=4$
when $\mathrm{p}=-1, \lambda_{2}=-4$
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 $\mathrm{y}^{2}$
\[
\Rightarrow \mathrm{k}^{2}=3 \mathrm{k}-2 \Rightarrow \mathrm{k}^{2}-3 \mathrm{k}+2=0
\]
\[
\Rightarrow(\mathrm{k}-1)(\mathrm{k}-2)=0 \Rightarrow \mathrm{k}=1,2
\]
(1) $\cap(2)$
$\mathrm{k}=1$ only Answer(2)
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