>$$ \vec b=\hat i+\hat j+\hat k $$

So,

$$ \vec b=(1,1,1) $$

For the plane

$$ \vec r\cdot(2\hat i+p\hat j+\hat k)=8 $$

the normal vector is

$$ \vec n=2\hat i+p\hat j+\hat k $$

So,

$$ \vec n=(2,p,1) $$

If $\theta$ is the acute angle between the line and the plane, then

$$ \sin\theta=\frac{|\vec b\cdot \vec n|}{|\vec b||\vec n|} $$

Here,

$$ \vec b\cdot \vec n=(1)(2)+(1)(p)+(1)(1)=p+3 $$

Also,

$$ |\vec b|=\sqrt{1^2+1^2+1^2}=\sqrt3 $$

and

$$ |\vec n|=\sqrt{2^2+p^2+1^2}=\sqrt{p^2+5} $$

Given,

$$ \theta=\sin^{-1}\left(\frac{\sqrt2}{3}\right) $$

So,

$$ \sin\theta=\frac{\sqrt2}{3} $$

Hence,

$$ \frac{|p+3|}{\sqrt3\sqrt{p^2+5}}=\frac{\sqrt2}{3} $$

Squaring both sides,

$$ \frac{(p+3)^2}{3(p^2+5)}=\frac{2}{9} $$

Cross-multiplying,

$$ 9(p+3)^2=6(p^2+5) $$

Divide by $3$:

$$ 3(p+3)^2=2(p^2+5) $$

Now expand:

$$ 3(p^2+6p+9)=2p^2+10 $$

$$ 3p^2+18p+27=2p^2+10 $$

$$ p^2+18p+17=0 $$

Factorising,

$$ p^2+18p+17=(p+1)(p+17) $$

So,

$$ (p+1)(p+17)=0 $$

Hence,

$$ p=-1 \quad \text{or} \quad p=-17 $$

So the correct option is

$$ \boxed{\text{Option B: } p=-1 \text{ or } p=-17} $$

Three Dimensional Geometry

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