Solution :

Condition for an equation to represent a pair of straight lines is

$abc+2fgh -af^2-bg^2-ch^2=0 $ --------(1)

Given $3x^2-11xy+10y^2-7x+13y+k=0$------(2)

Comparing equation (1) and (2)

$a=3,b=10,c=k$

$h= \large\frac{-11}{2}$

$g= \large\frac{-7}{2}$

$f= \large\frac{13}{2}$

Substituting the values we get,

$3 \times 10 \times k +2 \times \large\frac{13}{2} $$\times \bigg( \large\frac{-7}{2}\bigg) \times \bigg( \large\frac{-11}{2}\bigg) $$-3 \bigg( \large\frac{13}{2}\bigg)^2-10 \bigg( \large\frac{-7}{2}\bigg)^2- k \bigg( \large\frac{-11}{2}\bigg)^2=0$

=> $30 k +\bigg(\large\frac{1001}{4}\bigg) -\frac{507}{4}-\frac{490}{4} -\frac{121 \;k}{4}$

=> $\large\frac{120k +1001-507-490-121\;k}{4}$$=0$

$-k+4=0$

Hence $k=4$