The line joining the points \(6 \overrightarrow{\mathbf{a}}-4 \overrightarrow{\mathbf{b}}+4 \overrightarrow{\mathbf{c}},-4 \overrightarrow{\mathbf{c}}\) and the line joining the
points \(-\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}-3 \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-5 \overrightarrow{\mathbf{c}}\) intersect
at

The equations of the lines joining \(6 \overrightarrow{\mathbf{a}}-4 \overrightarrow{\mathbf{b}}+4 \overrightarrow{\mathbf{c}},-4 \overrightarrow{\mathbf{c}} \quad\) and
\(-\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}-3 \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-5 \overrightarrow{\mathbf{c}}\) are respectively
\(
\overrightarrow{\mathbf{r}}=6 \overrightarrow{\mathbf{a}}-4 \overrightarrow{\mathbf{b}}+4 \overrightarrow{\mathbf{c}}+m(-6 \overrightarrow{\mathbf{a}}+4 \overrightarrow{\mathbf{b}}-8 \overrightarrow{\mathbf{c}}) ...(i)
\)
and \(\overrightarrow{\mathbf{r}}=-\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}-3 \overrightarrow{\mathbf{c}}+n(2 \overrightarrow{\mathbf{a}}+4 \overrightarrow{\mathbf{b}}-2 \overrightarrow{\mathbf{c}})\) ...(ii)
For the point of intersection, the Eqs. (i) and
(ii) should give the same value of \(\overrightarrow{\mathbf{r}}\). Hence equating the coefficients of vectors \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) in the two expressions for \(\overrightarrow{\mathbf{r}}\), we get
\(
\begin{array}{l}
6 m+2 n=7 \\
2 m-2 n=1
\end{array}
\)
and
\(
8 m-2 n=7
\)
On solving Eqs. (iii) and (iv), we get \(m=1, n=\frac{1}{2} .\) These values of \(m\) and \(n\), also satisfy the Eq. (v).
\(\therefore\) The lines intersect. Putting the value of \(m\) in Eq. (i), we get the position vector of the point of intersection as \(-4 \overrightarrow{\mathbf{c}}\).