If an angle between the line, $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$
and the plane, $x-2 y-k z=3$ is $\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$,
then a value of $\mathrm{k}$ is:
Correct Option: , 3
DR's of line are $2,1,-2$
normal vector of plane is $\hat{i}-2 \hat{j}-k \hat{k}$
$\sin \alpha=\frac{(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\mathrm{k} \hat{\mathrm{k}})}{3 \sqrt{1+4+\mathrm{k}^{2}}}$
$\sin \alpha=\frac{2 \mathrm{k}}{3 \sqrt{\mathrm{k}^{2}+5}}$ .............(1)
$\cos \alpha=\frac{2 \sqrt{2}}{3}$ ...............(2)
$(1)^{2}+(2)^{2}=1 \Rightarrow k^{2}=\frac{5}{3}$
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