Let
${\bf A}$ = vector perpendicular to the first plane
${\bf B}$ = vector perpendicular to the second plane
${\bf A} = 2{\bf i} - 4{\bf j} - {\bf k}$
${\bf B} = 3{\bf i} + 4{\bf j} + 5{\bf k}$
The required angle between the two planes is equal to the angle between vectors $\mathbf{A}$ and $\mathbf{B}$. The formula for the angle between two vectors is given by
$\cos \theta = \dfrac{\mathbf{A} \cdot \mathbf{B}}{\left\| \mathbf{A} \right\| \, \left\| \mathbf{B} \right\| }$
Dot product of $\mathbf{A}$ and $\mathbf{B}$
$\mathbf{A} \cdot \mathbf{B} = 2(3) - 4(4) - 1(5)$
$\mathbf{A} \cdot \mathbf{B} = -15$
Magnitude of vector $\mathbf{A}$
$\left\| \mathbf{A} \right\| = \sqrt{2^2 + (-4)^2 + (-1)^2}$
$\left\| \mathbf{A} \right\| = \sqrt{21}$
Magnitude of vector $\mathbf{B}$
$\left\| \mathbf{B} \right\| = \sqrt{3^2 + 4^2 + 5^2}$
$\left\| \mathbf{B} \right\| = 5\sqrt{2}$
Hence,
$\cos \theta = \dfrac{-15}{\sqrt{21} \left( 5\sqrt{2} \right)}$
$\theta = 117.56^\circ$
$\text{Acute angle} = 180^\circ - \theta$
$\text{Acute angle} = 180^\circ - 117.56^\circ$
$\text{Acute angle} = 62.42^\circ$ ← answer
