Does Linear Cramer-Shoup have pseudo-random ciphertexts? "Linear Cramer-Shoup" is defined on pages 4 and 5 of $\:$ eprint.iacr.org/2007/074.pdf .

Are the ciphertexts in Linear Cramer-Shoup computationally indistinguishable from uniform 
under a chosen-ciphertext attack? $\:$ (We can assume without loss of generality either that

$M$ is the identity element or that $M$ is a a random element unknown to the adversary.)

This is not homework, and is relevant to a Zero Knowledge

Proof of Knowledge protocol I'm thinking about.
The security claim on page 5 of the Linear Cramer-Shoup paper is that their modified scheme is CCA secure, which is weaker than the IND-CCA2 security of the original DDH based Cramer-Shoup scheme. However, from the outline of the security proof, it seems the author actually means the LCS scheme is CCA2 secure. 

Also note the first sentence on page 6: 

"Clearly, if $A$ has a different advantage in guessing the bit $b$ when $B$ is run with a Linear tuple $(g_1,g_2,g_3,g_1^{r_1},g_2^{r_2},g_3^{r_1+r_2})$ and when $B$ is run with a random tuple $(g_1,g_2,g_3,g_1^{r_1},g_2^{r_2},\eta)$, we obtain a distinguisher for the Linear problem."

The cipher text $(u_1,u_2,u_3,e,v)$ consists of the last three elements of such a linear tuple, plus a blinded message $e$ and a verifier $v$. 

Since $u_1 = g_1^{r_1}$ and $u_2 = g_2^{r_2}$ and $r_1, r_2$ are selected at random, these two values are trivially indistinguishable from uniform, because they are uniform. Since $u_3 = g_3^{r_1 + r_2}$, this value is indistinguishable from uniform by the Linear assumption.

The blinded value $e$ is proven to be "independent of the view of the adversary", which I interpret as being indistinguishable from uniform. The proof is interesting. Basically, it is based on the following derived problem:

Given $(g^s,g^t,g^{sa},g^{tb},g^{a+b},g^{ad+be+cf})$ and any number of points on the "plane" $P(x,y) = g^{xd+ye+(x+y)f}$ except $P(a,b)$, output yes if $c = a + b$, no otherwise.

The uniformity of $v$ might be derived using the same technique. Set $g = g_3, g^z = g_1, g^w = g_2$. We then have:

$v^\prime = g^{r_1z(x_1+\alpha{y_1}) + r_2w(x_2+\alpha{y_2}) + r_3(x_3+\alpha{y_3})}$, where $v^\prime = v \iff r_3 = r_1 + r_2$.

Clearly, if $r_3$ is selected uniformly at random, $v^\prime$ is also uniform, so if you are unable to tell if $r_3 = r_1 + r_2$, you are unable to distinguish $v$ from uniform.