# If the angle $\dpi{100} \Theta$ between the line $\dpi{100} \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$  and the plane $\dpi{100} 2x-y+\sqrt{\lambda z}+4=0$ is such that $\dpi{100} \sin \Theta =\frac{1}{3},$ the value of $\dpi{100} \lambda$  is Option 1) $-\frac{3}{5}\;$ Option 2) $\; \frac{5}{3}\;$ Option 3) $\; \frac{-4}{3}\;$ Option 4) $\frac{3}{4}$

As we learnt in

Angle between line and Plane (vector form ) -

The angle between a line $\vec{r}= \vec{a}+\lambda \vec{b}$and the plane $\vec{r}.\vec{n}= d$ is given by

$\sin \Theta = \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}$

- wherein

Angle between line and normal to plane$= cos \left ( \frac{\pi}{2}-\theta \right )=\frac{ 2-1+2\sqrt{2}}{3\times \sqrt{5+\lambda }}$

$sin\theta =\frac{ 2\sqrt{\lambda }}{3 \sqrt{5+\lambda }}=\frac{1}{3}\, \, \, = > \lambda =\frac{5}{3}$

Option 1)

$-\frac{3}{5}\;$

Incorrect Option

Option 2)

$\; \frac{5}{3}\;$

Correct Option

Option 3)

$\; \frac{-4}{3}\;$

Incorrect Option

Option 4)

$\frac{3}{4}$

Incorrect Option

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