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Two planes $n_1\cdot r=d_1$, $n_2\cdot r=d_2$ with independent normal vectors meet in a straight line with direction $n_1\times n_2$. Any difference of two points on both planes is perpendicular to both normal vectors, which proves the direction statement. A particular point follows by solving the two linear equations; adding a real multiple of the direction gives the parametric equation of a straight line. For parallel normal vectors, the planes are either coincident or disjoint instead.

A plane through $r_0$ with nonzero normal vector $n$ consists of the position vectors $r$ satisfying $(r-r_0)\cdot n=0$. Thus $r\cdot n=d$, with $d=r_0\cdot n$. When $n$ is a unit vector, $d$ is the signed distance from the origin in the direction $n$; for an unnormalized normal vector, that distance is $d/\Vert n\Vert$. The displacement within the plane is perpendicular to $n$.

The cross product of the two normal vectors is $(2,6,-3)\times (5,2,-1)=(0,-13,-26)$, so their intersection of two planes has direction $(0,1,2)$. Eliminating $2y-z$ gives $x=2$, and substitution then gives $z=2y-2$. Thus the Cartesian constraints and parametric equation of a straight line areThere is no valid division by a zero $x$ component in a symmetric ratio form; retaining $x=2$ explicitly handles that component.

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

The straight line consists of $\mathbf{y}+s(\mathbf{z}-\mathbf{y})$, with $s\in \mathbb{R}$. Since the endpoints are distinct, the direction vector can be normalized. ChoosePutting $t=s|\mathbf{z}-\mathbf{y}|$ gives the required parametric equation of a straight line $\mathbf{r}=\mathbf{p}+t\hat{\mathbf{q}}$. Other choices of $\mathbf{p}$ on the same straight line, or the opposite direction for $\hat{\mathbf{q}}$, give equivalent descriptions.