Find the equations of the line passing through the point and parallel to the line . Also find the vector form of this equation so obtained.
Answer
Given: line passes through and is parallel to the line

$\frac{x–6}{3}=\frac{y–2}{–4}=\frac{z+7}{5}$
\frac{x-6}{3}=\frac{y-2}{-4}=\frac{z+7}{5}

To find: equation of line in vector and Cartesian form
Formula Used: Equation of a line is
Vector form:
Cartesian form:
where is a point on the line and is a vector parallel to the line.
Explanation:
Since the line say is parallel to another line say has the same direction ratios as that of
Here,
Since the equation of is

$\begin{array}{l}\frac{x–6}{3}=\frac{y–2}{–1}=\frac{z+7}{5}\\ \overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}–4\hat{\mathrm{l}}+5\hat{\mathrm{k}}\end{array}$
\begin{array}{l}
\frac{x-6}{3}=\frac{y-2}{-1}=\frac{z+7}{5} \\
\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{l}}+5 \hat{\mathrm{k}}
\end{array}

Therefore,
Vector form of the line is:

$\overrightarrow{r}=\hat{ı}–2\hat{ȷ}+3\hat{k}–\lambda –\lambda (3\hat{i}–4\hat{ȷ}+\sqrt{5}\hat{k})$
\vec{r}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k}-\lambda-\lambda(3 \hat{i}-4 \hat{\jmath}+\sqrt{5} \hat{k})

Cartesian form of the line is:

$\frac{x–1}{3}=\frac{y+2}{–4}=\frac{z–3}{5}$
\frac{x-1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}