A line is given by a pair of equations:
\begin{equation*}
a_1 x_1 +a_2 x_2+a_3 x_3 + a_4 x_4=0, \qquad
b_1 x_1 + b_2 x_2 + b_3 x_3 + b_4 x_4=0.
\end{equation*}
Suppose this line is on $X$.
If the minor $a_3 b_4-a_4 b_3$ is non-zero, then we may rewrite the equations of the line as
\begin{equation*}
x_3=a x_1+ b x_2, \qquad x_4=c x_1 + e x_2.
\end{equation*}
Substituting into the equation of the surface $X$ we see that the expression
\begin{equation*}
x_1^d+x_2^d-x_2^{d-2}(ax_1 +b x_2)(c x_1 + e x_2)
\end{equation*}
vanishes as polynomial in x_1 and x_2. This is clearly impossible by considering the coefficient of $x_1^d$. Hence the minor $a_3 b_4-a_4 b_3=0$. So we can suppose that one of the equations of the line is of the form $a x_1 + b x_2=0$. Assume that the line does not lie on either of the planes $x_1=0$ or $x_2=0$. Thus neither of $a$ or $b$ is zero and we may rewrite this equation as $x_1=c x_2$. Substituting in the equation for $X$ we see that the line lies on the conic
\begin{equation*}
(1+c^d) x^2 + x_2^2 - x_3 x_4=0.
\end{equation*}
If $1+c^d \ne 0$, then the conic is irreducible and so does not contain a line. Hence $1+c^d=0$ and so the line is on one of the planes $x_3=0$ or $x_4=0$.
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A line is given by a pair of equations:
\begin{equation*}
a_1 x_1 +a_2 x_2+a_3 x_3 + a_4 x_4=0, \qquad |
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