The area of triangle with vertices \((1,2,0),(1,0, a)\) and \((0,3,1)\) is \(\sqrt{6}\) sq. units, then the values of ' \(a\) ' are

Refer figure


\(
\mathrm{A}(\Delta \mathrm{ABC})=\frac{1}{2}|\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}|
\)
Here \(\quad \overline{B A}=2 \hat{j}-a \hat{k}\)
\(
\overline{\mathrm{BC}}=-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+(1-a) \hat{\mathrm{k}}
\)
Now \(\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 2 & -\mathrm{a} \\ -1 & 3 & 1-\mathrm{a}\end{array}\right|\)
\(
\begin{aligned}
& =\hat{i}(2-2 a+3 a)-\hat{j}(-a)+\hat{k}(2)=(a+2) \hat{i}+(a) \hat{j}+2 \hat{k} \\
\therefore & |\overline{B A} \times \overline{B C}|=\sqrt{(a+2)^2+(a)^2+(2)^2}
\end{aligned}
\)
From given data, we write
\(
\begin{aligned}
& \sqrt{6}=\frac{1}{2} \sqrt{2 a^2+4 a+8} \\
& \therefore 4(6)=2 a^2+4 a+8 \Rightarrow a^2+2 a-8 \\
& \Rightarrow(a+4)(a-2)=0 \\
& \Rightarrow a=-4,2
\end{aligned}
\)