Given,
$\begin{aligned}
A & =(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
B & =(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
C & =\frac{1}{4}(7 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}) \\
D & =\frac{1}{3}(7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+(5+3 a) \hat{\mathbf{k}}) \\
\mathbf{A C} & =\left(\frac{7}{4} \hat{\mathbf{i}}+\frac{15}{4} \hat{\mathbf{j}}+\frac{15}{4} \hat{\mathbf{k}}\right)-(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
\mathbf{A C} & =\frac{3}{4} \hat{\mathbf{i}}+\frac{7}{4} \hat{\mathbf{j}}+\frac{3}{4} \hat{\mathbf{k}} \\
\text {and } \mathbf{B D} & =\left(\frac{7}{3} \hat{\mathbf{i}}+\frac{2}{3} \hat{\mathbf{j}}+\left(\frac{5+3 a}{3}\right) \hat{\mathbf{k}}\right)-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
\mathbf{B D} & =\frac{1}{3} \hat{\mathbf{i}}+\frac{5}{3} \hat{\mathbf{j}}+\left(\frac{3 a-1}{3}\right) \hat{\mathbf{k}}
\end{aligned}$
Also given, $\quad|\mathbf{A C}|=|\mathbf{B D}|$
$\begin{aligned}
& \because\left(\frac{3}{4}\right)^2+\left(\frac{7}{4}\right)^2+\left(\frac{3}{4}\right)^2=\left(\frac{1}{3}\right)^2+\left(\frac{5}{3}\right)^2+\left(\frac{3 a-1}{3}\right)^2 \\
& \Rightarrow \quad \frac{9+49+9}{16}=\frac{1+25+(3 a-1)^2}{9} \\
& \Rightarrow \frac{67}{16}=\frac{26+(3 a-1)^2}{9}
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad 603=416+16(3 a-1)^2 \\
& \Rightarrow \quad 16(3 a-1)^2=187
\end{aligned}$