If the acute angle that the vector $α\hat i +β\hat j+γ\hat k$ makes with the plane of the two vectors $2\hat i+3\hat j-\hat k$ and $\hat i - \hat j + 2\hat k$ is $\tan^{-1}(\frac{1}{\sqrt{2}})$, then

$α (β+ γ) = βγ$

Let O be the angle between $\vec r = α\hat i +β\hat j+γ\hat k$ and the plane containing the vectors $\vec a = 2\hat i+3\hat j-\hat k$ and $\vec b=\hat i - \hat j + 2\hat k$. Then, $(π/2-θ)$ is the angle between $\vec r$ and $\vec a×\vec b$.

$∴\cos(\frac{π}{2}-θ)=\frac{\vec r.(\vec a×\vec b)}{|\vec r||\vec a×\vec b|}$

$⇒\sin θ=\frac{[\vec r\,\,\vec a\,\,\vec b]}{|\vec r||\vec a×\vec b|}$

$⇒\frac{1}{\sqrt{3}}=\frac{[\vec r\,\,\vec a\,\,\vec b]}{|\vec r|(5\sqrt{3})}$  $[∴\tan θ=\frac{1}{\sqrt{2}}⇒\sin θ=\frac{1}{\sqrt{3}}\,and\,\vec a×\vec b=5\hat i-5\hat j-5\hat k]$

$⇒5|\vec r|=[\vec r\,\,\vec a\,\,\vec b]$

$⇒5|\vec r|=5(α-β-γ)$

$⇒|\vec r|=α-β-γ$

$⇒\sqrt{α^2+β^2+γ^2}=α-β-γ$

$⇒βγ=αγ+αβ⇒βγ=α (β+ γ)$