Find the symmetric equations $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{6}\mathbf{)}\mathbf{}\mathbf{or}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{7}\mathbf{)}$and the parametric equations $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{)}$of a line, and/or the equation $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{10}\mathbf{)}$of the plane satisfying the following given conditions.

Line through $\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}$and parallel to the line $\mathbf{r}\mathbf{=}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{i}\mathbf{-}\mathbf{2}\mathbf{j}\mathbf{+}\mathbf{k}\mathbf{)}\mathbf{t}\mathbf{.}$

The symmetric equations of the line passing through $\mathbf{(}{\mathbf{x}}_{\mathbf{0}}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{,}{\mathbf{z}}_{\mathbf{0}}\mathbf{)}$and parallel to$\mathbf{ai}\mathbf{+}\mathbf{bj}\mathbf{+}\mathbf{ck}$ is $\frac{\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}}{\mathbf{a}}\mathbf{=}\frac{\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{0}}}{\mathbf{b}}\mathbf{=}\frac{\mathbf{z}\mathbf{-}{\mathbf{z}}_{\mathbf{0}}}{\mathbf{c}}\mathbf{;}$

The parametric equations of the line are $\mathbf{r}\mathbf{=}\mathbf{(}{\mathbf{x}}_{\mathbf{0}}\mathbf{+}\mathbf{at}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{+}\mathbf{bt}\mathbf{,}{\mathbf{z}}_{\mathbf{0}}\mathbf{+}\mathbf{ct}\mathbf{)}$.

The given point is $(5,-4,2)$and the line is localid="1657444744008" $\mathrm{r}=\mathrm{i}-\mathrm{j}+(5\mathrm{i}-2\mathrm{j}+\mathrm{k})\mathrm{t}$.

The given is in the form of localid="1657444831234" $\mathrm{r}={\mathrm{r}}_0+\mathrm{At}.$So, we get

localid="1657444841062" $\mathrm{A}=\mathrm{ai}+\mathrm{bj}+\mathrm{ck}$

localid="1657444905467" $\Rightarrow \mathrm{a}=5,\mathrm{b}=-2,\mathrm{c}=1$

The symmetric equations of the line passing through localid="1657444920128" $(5,-4,2)$and parallel to localid="1657444912200" $\mathrm{r}=\mathrm{i}-\mathrm{j}+(5\mathrm{i}-2\mathrm{j}+\mathrm{k})\mathrm{t}$is given by

localid="1657444949999" $\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_0}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_0}{\mathrm{c}}$

localid="1657444956034" $\Rightarrow \frac{\mathrm{x}-5}{5}=\frac{\mathrm{y}-(-4)}{-2}=\frac{\mathrm{z}-2}{1}$

localid="1657444964343" $\Rightarrow \frac{\mathrm{x}-5}{5}=\frac{\mathrm{y}+4}{-2}=\frac{\mathrm{z}-2}{1}$

Thus, the required solution is localid="1657444970433" $\frac{x-5}{5}=\frac{y+4}{-2}=\frac{z-2}{1}.$

The parametric equations of the straight line passing through $(5,-4,2)$and parallel to $\mathrm{r}=\mathrm{i}-\mathrm{j}+(5\mathrm{i}-2\mathrm{j}+\mathrm{k})\mathrm{t}$is given by

$\mathrm{x}=5+5\mathrm{t}$

$\mathrm{y}=-4-2\mathrm{t}$

$\mathrm{z}=2+\mathrm{t}$

Or

$\mathrm{r}=(5\mathrm{i}-4\mathrm{j}+2\mathrm{k})+(5\mathrm{i}-2\mathrm{j}+\mathrm{k})\mathrm{t}.$

Thus, the required solution is$\mathrm{r}=(5\mathrm{i}-4\mathrm{j}+2\mathrm{k})+(5\mathrm{i}-2\mathrm{j}+\mathrm{k})\mathrm{t}.$