Given \(\overrightarrow{\mathbf{p}}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
\(\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{p}}=x \overrightarrow{\mathbf{a}}+y \overrightarrow{\mathbf{b}}+z \overrightarrow{\mathbf{c}}\), then \(x, y, z\)
are respectively

\(
\text { } \begin{aligned}
& \overrightarrow{\mathbf{p}}=x \overrightarrow{\mathbf{a}}+y \overrightarrow{\mathbf{b}}+z \overrightarrow{\mathbf{c}} \\
& \Rightarrow 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=x(\hat{\mathbf{i}}+\hat{\mathbf{j}})+y(\hat{\mathbf{j}}+\hat{\mathbf{k}})+z(\hat{\mathbf{i}}+\hat{\mathbf{k}}) \\
& \Rightarrow 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=(x+z) \hat{\mathbf{i}}+(x+y) \hat{\mathbf{j}}+(y+z) \hat{\mathbf{k}}
\end{aligned}
\)
On comparing both sides the coefficients of \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\), we get
\(
x+z=3
\)
\(
x+y=2 \quad \ldots
\)
\(
\text { and } \quad y+z=4 \quad \ldots \text { (iii) }
\)
On solving Eqs. (i), (ii) and (iii), we get
\(
x=\frac{1}{2}, \quad y=\frac{3}{2}, \quad z=\frac{5}{2}
\)