$ \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $
$(\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_1-\overline{\mathrm{a}}_2$
$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$ \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} $
$ \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} $
$ 132=|6 \lambda+126| $
$ \lambda=1, \lambda=-43 $
$ |\lambda|=43$
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$\overline{A B}=-2 \hat{i}+\hat{j}+3 \hat{k}$
$\overline{C B}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$
$\overline{C A}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$
If $\delta > 0$ and the area of the triangle $ABC$ is $5 \sqrt{6}$, then $\overline{C B} \cdot \overline{C A}$ is equal to
