Given data

$\vec a = \hat i + \alpha \hat j + \beta \hat k$ so $|\vec a|^2 = 1+\alpha^2+\beta^2.$

Also given:

• angle between $\vec a$ and $\vec b$ is $\dfrac{\pi}{4}$,
• $|\vec b|^2=6 \Rightarrow |\vec b|=\sqrt{6}$,
• $\vec a\cdot \vec b = 3\sqrt{2}$.

Use the dot product formula

$\vec a\cdot \vec b = |\vec a||\vec b|\cos\frac{\pi}{4}$

Substitute the values: $3\sqrt{2} = |\vec a|\cdot \sqrt{6}\cdot \frac{1}{\sqrt{2}}$

$3\sqrt{2} = |\vec a|\sqrt{3}$

Hence, $|\vec a| = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{6}.$

Therefore, $|\vec a|^2 = 6.$

Find $\alpha^2+\beta^2$

Since $|\vec a|^2 = 1+\alpha^2+\beta^2,$ we get $1+\alpha^2+\beta^2 = 6$ $\alpha^2+\beta^2 = 5.$

Find $|\vec a\times \vec b|^2$

Using $|\vec a\times \vec b| = |\vec a||\vec b|\sin\frac{\pi}{4},$ we get $|\vec a\times \vec b|^2 = |\vec a|^2|\vec b|^2\sin^2\frac{\pi}{4}.$

Substitute values: $|\vec a\times \vec b|^2 = 6\cdot 6\cdot \frac{1}{2} = 18.$

Compute the required expression

$\left(\alpha^2+\beta^2\right)|\vec a\times \vec b|^2 = 5\cdot 18 = 90.$

Check options

The correct option is: $\boxed{\text{B: }90}$