$\begin{aligned} \vec{\alpha} &=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \vec{\beta} &=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \vec{\gamma} &=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}} \end{aligned}$
Let $\vec{\delta}=a \hat{i}+b \hat{j}+c \hat{k}$
Since $\vec{\alpha}$ and $\vec{\delta}$ are perpendicular, $\vec{\alpha} \cdot \vec{\delta}=0$
$\Rightarrow \mathrm{a}+2 \mathrm{~b}-\mathrm{c}=0$
$\vec{\beta}$ and $\vec{\delta}$ are perpendicular, $\vec{\beta} \cdot \vec{\delta}=0$
$\Rightarrow 2 \mathrm{a}-\mathrm{b}+3 \mathrm{c}=0$
from $(1),(2), \frac{a}{5}=\frac{b}{-5}=\frac{c}{-5} \Rightarrow \frac{a}{1}=\frac{b}{-1}=\frac{c}{-1}=x($ say $)$
So, $a=x, b=-x, c=-x$
Also, it is given $\vec{\gamma} \cdot \vec{\delta}=10$
$\Rightarrow 2 \mathrm{a}+\mathrm{b}+6 \mathrm{c}=10$
$\Rightarrow 2 x-x-6 x=10$
$\Rightarrow-5 x=10$
$\Rightarrow x=-2$
$\therefore \vec{\delta}=-2 \hat{i}+2 \hat{j}+2 \hat{k} .$
$\therefore|\vec{\delta}|=\sqrt{4+4+4}=\sqrt{12}=2 \sqrt{3}$