ABCD is a rhombus. If $\overline{\mathrm{AC}}=\overline{\mathrm{i}}+(1+\mathrm{\lambda })\overline{\mathrm{j}}+(\mathrm{\lambda }-2)\overline{\mathrm{k}}\text{ and }\overline{\mathrm{BD}}=(2\mathrm{\lambda }-1)\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\text{ then }\mathrm{\lambda }=$

$$\begin{gathered} In rhombus diagonals perpendicular to each other \\ \overline{\mathrm{AC}}\cdot \overline{\mathrm{BD}}=0 \\ \Rightarrow 1(2\lambda -1)+(1+\lambda )\left(1\right)+(\lambda -2)(-1)=0 \\ \Rightarrow 2\lambda -1+1+\lambda -\lambda +2=0 \\ \Rightarrow 2\lambda =-2 \\ \Rightarrow \lambda =-1 \end{gathered}$$