Three planes kx + y + z = 2, x + y – z = 3, x + 2z = 2 form a triangular prism and area of the normal section (where normal section of the triangular prism means the planar section which is parallel to the triangular base of the prism) be t. Then the value of  $5\sqrt{14}\left(kt\right)$ is___________

For triangular prism  $\left|\begin{array}{ccc}4& 1& 1\\ 1& 1& -1\\ 1& 0& 2\end{array}\right|=0$
$2k-3+(-1)=0\Rightarrow k=2$ 
The three planes are  $2x+y+z=2,x+y-z=3,x+2z-2=0$
Equation of plane passing through the line of intersection of first two planes is $(2x+y+z-2)+\lambda (x+y-z-3)=0$

$(2+\lambda )x+(1+\lambda )y+(1-\lambda )z-(2+3\lambda )=0$
equation of plane part of this family and parallel to third plane is $x+2z+1=0$

perpendicular distance between planes $=\frac{3}{\sqrt{5}}$

$$\begin{gathered} \sin \theta =\frac{\sqrt{14}}{\sqrt{30}},\sin \varphi =\frac{\sqrt{14}}{\sqrt{15}} \\ AB=h/\sin \theta =\frac{3\sqrt{30}}{\sqrt{75}} \\ AC=\frac{h}{\sin \varphi }=\frac{3\sqrt{15}}{\sqrt{70}} \\ \sin \varphi =\frac{\sqrt{14}}{3\sqrt{2}} \end{gathered}$$

Area of the triangle  $=\frac{1}{2}\left(AB\right)\left(AC\right)\sin \varphi$

$$\begin{gathered} =\frac{1}{2}\left(\frac{3\sqrt{30}}{\sqrt{75}}\right)\left(\frac{3\sqrt{15}}{\sqrt{70}}\right)\left(\frac{\sqrt{14}}{3\sqrt{2}}\right) \\ t=\frac{9}{2\sqrt{14}} \\ 5\sqrt{14}k{t}_1=\left(5\sqrt{14}\right)\left(2\right)\left(\frac{9}{2\sqrt{14}}\right) \\ = \left(5\right) \left(9\right) = 45 \end{gathered}$$