To prove that $B(t) = (B_1(t), B_2(t))$ is a 2-dimensional Brownian motion, we need to ensure that $B_1(t)$ and $B_2(t)$ satisfy the requirements of Brownian motion. Specifically:

1. Each $B_i(t)$ must be normally distributed with mean zero and variance $t$.
2. The increments of $B_i(t)$ are independent and stationary.
3. $B_1(t)$ and $B_2(t)$ must be uncorrelated.

Given: $B_1(t) = \gamma W_1(t) + \alpha W_2(t) - \sqrt{11} W_3(t)$ $B_2(t) = 2 W_1(t) + \beta W_2(t) + 3\sqrt{2} W_3(t)$

we proceed as follows:

Step 1: Variance Condition

We need the variances of $B_1(t)$ and $B_2(t)$ to each equal $t$. To find this, we calculate the variances of $B_1(t)$ and $B_2(t)$ based on the variances of the components $W_1(t)$, $W_2(t)$, and $W_3(t)$, each of which is an independent 1-dimensional Brownian motion with variance $t$. Thus: $\operatorname{Var}(W_1(t)) = \operatorname{Var}(W_2(t)) = \operatorname{Var}(W_3(t)) = t.$

For $B_1(t)$:

= \gamma^2 t + \alpha^2 t + 11t = t.$$ Dividing both sides by $$t$$, we get: $$\gamma^2 + \alpha^2 + 11 = 1.$$ For $$B_2(t)$$: $$\operatorname{Var}(B_2(t)) = 4 \operatorname{Var}(W_1(t)) + \beta^2 \operatorname{Var}(W_2(t)) + (3\sqrt{2})^2 \operatorname{Var}(W_3(t)) = 4t + \beta^2 t + 18t = t.$$ Dividing by $$t$$: $$4 + \beta^2 + 18 = 1,$$ which simplifies to: $$\beta^2 = -21.$$