Find the constant k so that the planes x–2y+kz=0 and 2x + 5y – z = 0 are at right angles.

Given planes x - 2y + kz = 0 .... (1)

compare with $a_{1}x+b_{1}y+c_{1}z+d_1=0$
$$\begin{gathered} \mathrm{Where} {\mathrm{a}}_1=1,b_1=-2,{\mathrm{c}}_1=\mathrm{k} \\ 2\mathrm{x}+5\mathrm{y}-\mathrm{z}=0 ...\left(2\right) \\ compare with {\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\mathrm{z}+{\mathrm{d}}_2=0 \\ Where \begin{array}{l}{\mathrm{a}}_2=2,\hspace{0.25em}{\mathrm{b}}_2=5\hspace{0.25em},\hspace{0.25em}{\mathrm{c}}_2=-1\end{array} \end{gathered}$$
Planes (1) & (2) are perpendicular then $\cos \mathrm{\theta }=0$
$\begin{array}{l}\Rightarrow {\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=0\Rightarrow 1\left(2\right)+(-2)5+\mathrm{k}(-1)=0\\ \Rightarrow 2-10-\mathrm{k}=0\\ \Rightarrow -8-\mathrm{k}=0\\ \therefore \mathrm{k}=-8\end{array}$