$\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
(A) $\mathbf{a}-\mathbf{b}=\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
Then, unit vector is
$\frac{-\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{\sqrt{1+36+36}}=\frac{-\hat{\mathbf{i}}}{\sqrt{73}}-\frac{6 \hat{\mathbf{j}}}{\sqrt{73}}-\frac{6 \hat{\mathbf{k}}}{\sqrt{3}}$
(B) $\mathbf{A B}=\mathbf{a}=\mathbf{B C}=\mathbf{b}$, then $\mathbf{C A}=\mathbf{c}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{k}}$
or $-3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}$ ...(iii)
(C) Centroid is

$\begin{aligned}
& =\frac{(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})+(3 \hat{\mathbf{i}}-4 \hat{\mathbf{k}})}{3} \\
& =\frac{6 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}}{3}=2 \hat{\mathbf{i}}-\frac{8}{3} \hat{\mathbf{k}}
\end{aligned}$
(D) Vector parallel to a
$\begin{aligned}
& =k(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
\text { Magnitude } & =k \sqrt{4+9+1} \\
\Rightarrow \quad \sqrt{14} k & =2 \sqrt{14} \\
k & =2 \\
\text {So, } \quad d & =4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\text {Now, } \quad \mathbf{b}-\mathbf{d} & =3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}
\end{aligned}$
So,
(a) $\rightarrow$ (iv);
(b) $\rightarrow$ (iii);
(c) $\rightarrow$ (ii);
(d) $\rightarrow$ (i)