Given:

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}$

$\overrightarrow{c}=\hat{i}+3\hat{j}-2\hat{k}$

$\overrightarrow{d}=2\hat{i}+\hat{j}-\hat{k}$

Also,

$l=\overrightarrow{b}·\overrightarrow{c}$ 

$\Rightarrow l=\left(\hat{i}-2\hat{j}+\hat{k}\right)·\left(\hat{i}+3\hat{j}-2\hat{k}\right)$

$\Rightarrow l=-7$

And,

$m=\overrightarrow{b}·\overrightarrow{a}$

$\Rightarrow m=\left(\hat{i}-2\hat{j}+\hat{k}\right)·\left(\hat{i}+\hat{j}+\hat{k}\right)$

$\Rightarrow m=0$

We have,

$\left[\begin{array}{ccc}\left(m\overrightarrow{b}+I\overrightarrow{a}\right)& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]$

$=\left[\begin{array}{ccc}-7\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]$

$=\left[\begin{array}{ccc}-7\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]$

$=-7\left(\hat{i}+\hat{j}+\hat{k}\right)·\left[\left(\hat{i}-2\hat{j}+\hat{k}\right)\times \left(2\hat{i}+\hat{j}-\hat{k}\right)\right]$

$=-7\left(\hat{i}+\hat{j}+\hat{k}\right)·\left(\hat{i}+3\hat{j}+5\hat{k}\right)$

$=-7\times \left(1+3+5\right)$

$=-63$