The area of the parallelogram with vertices \(\mathrm{A}(1,2,3), \mathrm{B}(1,3, \mathrm{a})\), \(C(3,8,6)\) and \(D(3,7,3)\) is \(\sqrt{265}\) sq. units, then \(a=\)

Refer figure

\(
\begin{gathered}
\mathrm{A}(\triangle \mathrm{ABC})=\frac{\mathrm{A}(\square \mathrm{ABCD})}{2} \\
=\frac{\sqrt{265}}{2}
\end{gathered}
\)
Also \(\mathrm{A}(\triangle \mathrm{ABC})=\frac{1}{2}|\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}|\)
Here \(\overline{\mathrm{BA}}=-\hat{\mathrm{j}}+(3-\mathrm{a}) \hat{\mathrm{k}}\) and \(\overline{\mathrm{BC}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+(6-\mathrm{a}) \hat{\mathrm{k}}\)
\(
\begin{aligned}
& \text { Now } \overline{B A} \times \overline{B C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & -1 & (3-a) \\
2 & 5 & (6-a)
\end{array}\right| \\
& =\hat{i}(-6+a-15+5 a)-\hat{j}(-6+2 a)+\hat{k}(2) \\
& =(6 a-21) \hat{i}-(2 a-6) \hat{j}+2 \hat{k}
\end{aligned}
\)
Magnitude of \(\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}=\sqrt{(6 \mathrm{a}-21)^2+(2 \mathrm{a}-6)^2+4}\)
\(
\begin{aligned}
& \therefore \frac{1}{2} \sqrt{(6 a-21)^2+(2 a-6)^2+4}=\frac{\sqrt{265}}{2} \\
& \therefore(6 a-21)^2+(2 a-6)^2+4=265 \\
& \therefore 40 a^2-276 a+216=0
\end{aligned}
\)

\(\therefore a=6\) from option given.