Multivariate random vector with normal distribution Let $X=(X_1, X_2)'\in N(\mu, \Lambda),$ where 
$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_1=3).$
Don't know even where to start. 
 A: Currently your question says

$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_3=3).$

I will assume you just forgot $X_3.$ You have
$$
\begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} \sim N\left( \begin{bmatrix} 1 \\ 1 \\ \mu_3 \end{bmatrix} , \begin{bmatrix} 3 & 1 & \sigma_{1,3} \\ 1 & 2 & \sigma_{2,3} \\ \sigma_{1,3} & \sigma_{2,3} & \sigma_{3,3} \end{bmatrix} \right).
$$
Let $Y = X_2+3X_3.$ Then
$$
\begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix},
$$
so that
$$
\operatorname{E} \begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \mu_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 + 3\mu_3 \end{bmatrix} 
$$
and
$$
\operatorname{var} \begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 & \sigma_{1,3} \\ 1 & 2 & \sigma_{2,3} \\ \sigma_{1,3} & \sigma_{2,3} & \sigma_{3,3} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1+3\sigma_{1,3} \\ 1+3\sigma_{1,3} & 2+6\sigma_{2,3} + 9\sigma_{3,3}  \end{bmatrix}.
$$
Find that bivariate normal density and plug in $3$ in place of the second argument, for $Y=3.$ Don't worry about the normalizing constant; you can read of the expected value and the variance from other aspects of the form of the density function.
Or alternatively, use one of the usual formulas for the conditional distribution of one component of a bivariate normal random variable given the value of the other component.
