The following planes are parallel. \( \begin{array}{l}4 x+6 y+5 z+14=0 \text { and } 16 x+24 y+20 z=0 \\ \text { Find the distance between the planes. } \\ d=\text { Ex: } 5.67\end{array} \)

To find the distance between the given parallel planes, we can use the formula: \[ d = \frac{|d_2 - d_1|}{\sqrt{A^2 + B^2 + C^2}} \] First, we need to rewrite the equations of the planes in the form \(Ax + By + Cz + D = 0\). The first plane is already in the correct form: \[ 4x + 6y + 5z + 14 = 0 \] Here, \(D_1 = 14\), and the coefficients are \(A_1 = 4\), \(B_1 = 6\), \(C_1 = 5\). The second plane is given by: \[ 16x + 24y + 20z = 0 \] This can be rewritten as: \[ 16x + 24y + 20z + 0 = 0 \] Here, \(D_2 = 0\), and the coefficients are \(A_2 = 16\), \(B_2 = 24\), \(C_2 = 20\). Next, we can observe that the second plane can be simplified as it is a multiple of the first plane: \[ \left(4x + 6y + 5z + 14\right) \times 4 = 16x + 24y + 20z + 56 = 0 \quad \Rightarrow \quad 16x + 24y + 20z = -56 \] So now, we can find the distance using the formula. The coefficients \(A\), \(B\), and \(C\) remain the same across both planes since they are parallel. We calculate: \[ d = \frac{|0 - 14|}{\sqrt{4^2 + 6^2 + 5^2}} = \frac{14}{\sqrt{16 + 36 + 25}} = \frac{14}{\sqrt{77}} \] Calculating \( \sqrt{77}\) gives approximately \(8.77\), thus: \[ d \approx \frac{14}{8.77} \approx 1.59 \] So, the distance \(d\) between the planes is approximately: \(\boxed{1.59}\)