Finding a critical region for a mixture Let $\ X_1 , X_2 $ be two iid random variables with normal N ( $\theta,1 $) distribution.  Further , consider bernoulli random variable V with P(V=1) = $ \frac{1}{4} $ and which is independent of $\ X_1 , X_2 $ .
Define $ \ X_3 $ as =
$\ X_1 $, if V=0
$\ X_2 $, if V=1
For testing hypothesis H0 : $ \theta =0 $ versus H1 : $ \theta =1 $
Reject H0 if $\frac{\ X_1 + X_2 +X_3 }{3} $ >C
Find C such that test size becomes 0.05 .
Now in this question I could understand only two things , one that V is an indicator variable and bernoulli distribution and second that it should be converted to Z before adding two critical region.
Other than that I dont know how to proceed , please help.
My attempt
$ \ X_3 = \ X_1*P(v=0) + \ X_2*P(v=1 ) $
$ E(\ X_3) = E(\ X_1*P(v=0)) + E(\ X_2*P(v=1 )) $
$ E(\ X_3) = (\theta *3/4) + (\theta*1/4) $
$ E(\ X_3) = \theta  $
Now $ Var (\ X_3) = Var(\ X_1 *3/4) +  var(\ X_2*1/4) $
$ Var (\ X_3) = \frac{9}{16} +  \frac{1}{16} $
$ Var (\ X_3) = \frac{10}{16}  $
$\frac{\ X_1 + X_2 +X_3 }{1} \ ~~~ N( 3\theta , ??) $
$\frac{\ X_1 + X_2 +X_3 }{3} \ ~~~ N( \theta , ??) $
Please help to proceed .

A mixture of identically distributed variables has the same distribution -- but is not independent of its component variables.
Some of the information in this exercise is just a distraction.  Stripped to its essential underlying idea, the problem is the following:

Let $X_1,X_2$ be independent identically distributed random variables.  Use an independent Bernoulli$(p)$ variable $V$ to define $X_3 = X_1$ when $V=0$ and otherwise $X_3=X_2.$  What is the distribution of $Y = X_1+X_2+X_3$?

The iid assumption implies $(X_1,X_2)$ and $(X_2,X_1)$ have the same distribution: that is, we may switch the roles of the $X_i.$  Upon doing this with the random variable $2X_1+X_2$ it is immediate that

$2X_1+X_2$ and $X_1 + 2X_2$ have the same distribution.

Let this common distribution function be $F.$  This means only that for any number $y,$
$$\Pr(2X_1+X_2 \le y) = F(y) = \Pr(X_1+2X_2 \le y.)$$
The distribution function of $Y$ is found by studying the two possibilities for $V,$ using the facts that when $V=0,$ $Y=2X_1+X_2$ and when $V=1,$ $Y=X_1+2X_2.$  The law of total probability asserts
$$\begin{aligned}
\Pr(Y\le y) &= \Pr(V=0)\Pr(2X_1 + X_2 \le y) + \Pr(V=1)\Pr(X_1+2X_2\le y)\\
& = (1-p)F(y) + pF(y) \\
&= F(y).
\end{aligned}$$
Thus,

$Y$ has the same distribution as $2X_1+X_2$ and $X_1 + 2X_2.$

The rest is mopping up: in your circumstance, where the $X_i$ have Normal distributions and are independent, $Y$ must therefore have a Normal distribution and its parameters are easily computed.  From that it's simple to find the value of $C$ in the question.

The joint distribution of all the relevant variables is unusual and instructive.  In this scatterplot matrix of 2000 realizations (with $\theta=5$) I have colored the points where $V=0$ in blue and those where $V=1$ in red.  Only $X_1$ and $X_2$ are independent, as indicated by their circular-cloud scatterplots.  The scatterplots with $Y$ include some singular parts on the diagonal, reflecting the fact that often $Y$ equals $2X_1+X_2$ or $X_1+2X_2.$

