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I know that there is a correspondence between points in $\mathbb{P}^5$ and conics in $\mathbb{P}^2$.

How do you show that family of conics through four points (pencil of conics) in general position forms a line when it's considered in $\mathbb{P}^5$?

If I consider a general conic given by $$ax^2+by^2+cxy+dx+ey+f=0$$ we can divide both sides by $a$ if $a$ is nonzero $$x^2+b'y^2+c'xy+d'x+e'y+f'=0,$$ and if I'm given the coordinates of four points, I can solve this for four variables, where the other one is a free variable. That's where the linearity comes from.

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You don't have to worry so much about whether $a$ is zero or nonzero: you can just write down what the conditions "passing through $p$" mean. Explicitly, if $p=[p_0,p_1,p_2]$, then a conic $ax^2+by^2+cxy+dxz+eyz+fz^2$ passes through $p$ iff $ap_0^2+bp_1^2+cp_0p_1+dp_0p_2+ep_1p_2+fp_2^2=0$, which is a linear equation in $a,b,c,d,e,f$. Assuming the points are in general position gives you $4$ independent linear equations, so all you need to do is to understand that $V(l_1,l_2,l_3,l_4)$ is a $\Bbb P^1\subset \Bbb P^5$. (I should note that your equation of a conic is an affine equation and I'm upgrading to a projective equation - you should too!)

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