Answer:

Use coordinates

\(A=(1,0,0),\qquad B=(0,2,0),\qquad C=(0,0,4).\)

Two vectors in the plane \(ABC\) are

\(\overrightarrow{AB}=(-1,2,0),\qquad \overrightarrow{AC}=(-1,0,4).\)

Their cross product is

\(\overrightarrow{AB}\times\overrightarrow{AC}=(8,4,2)=2(4,2,1).\)

So a vector perpendicular to the plane is \(4\mathbf{i}+2\mathbf{j}+\mathbf{k}\).

The plane \(ABC\) has equation

\(4x+2y+z=4.\)

The perpendicular distance from \(O\) to this plane is

\(ON=\frac{|4|}{\sqrt{4^{2}+2^{2}+1^{2}}}=\frac{4}{\sqrt{21}}.\)

The foot \(N\) lies on the line through \(O\) in the normal direction, so write

\(N=t(4,2,1).\)

Substitute into the plane equation:

\(4(4t)+2(2t)+t=21t=4,\)

so \(t=\frac{4}{21}\), and

\(N=\left(\frac{16}{21},\frac{8}{21},\frac{4}{21}\right).\)

Since \(OP=\frac{3}{4}ON\),

\(P=\frac34N=\left(\frac47,\frac27,\frac17\right).\)

The line \(AP\) can be written as

\((x,y,z)=(1,0,0)+s\left(-\frac37,\frac27,\frac17\right).\)

The plane \(OBC\) is \(x=0\). Hence

\(1-\frac37s=0\quad\Rightarrow\quad s=\frac73.\)

Therefore

\(Q=\left(0,\frac23,\frac13\right).\)

A normal vector to plane \(ABQ\) is obtained from

\(\overrightarrow{AB}=(-1,2,0),\qquad \overrightarrow{AQ}=\left(-1,\frac23,\frac13\right).\)

The cross product is proportional to

\((2,1,4).\)

The normals to \(ABC\) and \(ABQ\) may therefore be taken as \((4,2,1)\) and \((2,1,4)\). Thus

\(\cos\theta=\frac{(4,2,1)\cdot(2,1,4)}{\sqrt{4^{2}+2^{2}+1^{2}}\sqrt{2^{2}+1^{2}+4^{2}}}=\frac{14}{21}=\frac23.\)

So the acute angle is

\(\theta=\cos^{-1}\left(\frac23\right).\)