Question: I have two dependent Gaussian variables $X\_1,X\_2$ with unit mean each, and standard deviations $\sigma\_1=2a$ and $\sigma\_2=a$ respectively, while $a>0$

Is the joint distribution of these two dependent gaussian variables also gaussian !?

noting that their dependence is in the shape 
$$X\_2 = bX\_1+X\_3$$
where $X\_3$ is another Gaussian RV with unit mean and constant variance.
$X\_1,X\_3$ are independent

Please advise

Answer: 
The answer is yes. Note that two random variables $X, Y$ are joint Gaussian iff $aX + bY$ is Gaussian for all $a, b \in \mathbb{R}$.
Yes, in your case, the joint distribution of two Gaussian random variables will be Gaussian, but this is not generally true (as per the comments).

Using characteristic functions, one can show that $(X\_1,X\_2)$ is multivariate Gaussian, iff $a\_1X\_1+a\_2X\_2$ is univariate Gaussian for all choices of non-zero vectors $(a\_1,a\_2)\in\Bbb{R}^2\setminus\{0\}$.

Let $\Sigma$ be the covariance matrix of $(X\_1,X\_2)$, i.e. $\Sigma$ is the $2\times 2$-matrix, whose diagonal elements are the variances of $X\_1$ and $X\_2$, and whose off-diagonal elements are the covariance between $X\_1$ and $X\_2$.

Given that $X\_1$ and $X\_2$ are not fully correlated, the joint density of $(X\_1,X\_2)$ is
$$f\_{(X\_1,X\_2)}(\mathbf{x}) = \frac{1}{2\pi \sqrt{|\Sigma|}}\exp\left(-\frac{1}{2}(\mathbf{x}-\boldsymbol{\mu})^T\Sigma^{-1}(\mathbf{x}-\boldsymbol{\mu})\right),$$
where $\mathbf{x} = (x\_1,x\_2)$, $\boldsymbol{\mu} = (\mu\_1,\mu\_2)$ are the mean values of $(X\_1,X\_2)$, and where $|\Sigma|$ is the determinant of $\Sigma$.