$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\mathrm{p} \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$
$\because \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \& \overrightarrow{\mathrm{c}}$ are coplanar
$$
\begin{aligned}
& \therefore a(\vec{b} \times \vec{c})=[\vec{a}, \vec{b}, \vec{c}]=0 \\
& \Rightarrow\left|\begin{array}{ccc}
2 & 3 & -1 \\
4 & -1 & 3 \\
p & 1 & -1
\end{array}\right| \\
& \Rightarrow 2(1-3)-3(-4-3 p)-1(4+p)=0 \\
& \Rightarrow-4+12+9 p-4-p=0 \\
& \Rightarrow 8 p=-6 \Rightarrow p=\frac{-3}{4} \Rightarrow 8 p=-4 \Rightarrow p=-\frac{1}{2}
\end{aligned}
$$
$\begin{aligned} & \text { Now, } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & -1 \\ -\frac{1}{2} & 1 & -1\end{array}\right| \\ & \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}+\frac{5}{2} \hat{\mathrm{j}}+\frac{7}{2} \hat{\mathrm{k}} \\ & \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=\sqrt{4+\frac{25}{4}+\frac{49}{4}}=\sqrt{\frac{90}{4}}=\frac{3 \sqrt{10}}{2}\end{aligned}$