Does this set of vectors span the space of all polynomials of degree at most 3?

Since $P_3$ has dimension 4, the set $\{p_1,p_2,p_3,p_4\}$ will span $P_3$ if and only if they are independent. So, test if the vectors in $\{p_1,p_2,p_3,p_4\}$ are independent:

Assume $$ c_1p_1+c_2p_2+c_3p_3+c_4p_4={\bf 0}. $$ Then $$ c_1(x^3-2x^2+x+1)+c_2(x^2-x+2)+c_3(2x^3+3x+4)+c_4(3x^2+2x+1) ={\bf 0}. $$ Collecting like terms, the above can be written as $$ (c_1+2c_3)x^3+(-2c_1+c_2+3c_4)x^2+(c_1-c_2+3c_3+2c_4)x+(c_1+2c_2+4c_3+c_4)={\bf 0}. $$ A polynomial is the zero polynomial if and only if all its coefficients are 0; so, the above is equivalent to the following system of equations: $$\tag{1}\eqalign{ c_1+ 2c_3 &=0\cr -2c_1+c_2+3c_4&=0\cr c_1-c_2+3c_3+2c_4&=0\cr c_1+2c_2+4c_3+c_4&=0} $$ The coefficient matrix of the above system is $$ A=\left[\matrix{1&0&2&0\cr -2&1&0&3\cr 1&-1&3&2\cr 1&2&4&1\cr }\right] $$ An echelon form of $A$ is $$ \left[\matrix{1&0&2&0\cr 0&1&-4&3\cr 0&0&-3&5\cr 0&0&0&7\cr }\right]. $$ This implies the system above has only the trivial solution: $c_1=c_2=c_3=c_4=0$; thus $\{p_1,p_2,p_3,p_4\}$ is an independent set and so spans $P_3$.

Alternatively you could show that the equation you wrote always has a solution. To do this, write the corresponding system of equations. You'll wind up with the system as in $(1)$, but with the right hand side replace by the coefficients of $y$.

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If you just want to "cut to the chase", note that the coefficient matrix $A$ above is simply the matrix whose columns are the coefficients of the polynomials $p_1$, $p_2$, $p_3$, $p_4$. You could have immediately written this down (or written them down as rows) and then determined if the matrix has full rank. It's a good thing to see why you can do this, though...