If the angle $ heta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ is such that $\sin heta=\frac{1}{3}$, then $\lambda+1=$

Question

If the angle $	heta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ is such that $\sin 	heta=\frac{1}{3}$, then $\lambda+1=$

Accepted Answer

Correct answer: C

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