Find the parametric form of the line that goes through the point $$ A=(2,-1,5) $$ and is parallel to the vector $$ \mathbf{v}=[-4,-3,3] $$. $$ \begin{array}{l}x=\text { Ex: } 6+8 \mathrm{t} \ y=\square \ z=\square\end{array} $$

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The general parametric form of a line is given by
$$
\mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v},
$$
where $$\mathbf{r}_0$$ is a point on the line and $$\mathbf{v}$$ is a direction vector.

Given:
- Point $$ A = (2, -1, 5) $$. So, $$\mathbf{r}_0 = \begin{pmatrix} 2 \ -1 \ 5 \end{pmatrix}$$.
- Direction vector $$ \mathbf{v} = \begin{pmatrix} -4 \ -3 \ 3 \end{pmatrix}$$.

Substituting these into the parametric form:
$$
\mathbf{r}(t) = \begin{pmatrix} 2 \ -1 \ 5 \end{pmatrix} + t \begin{pmatrix} -4 \ -3 \ 3 \end{pmatrix}.
$$

This gives the component form:
$$
\begin{aligned}
x &= 2 - 4t, \
y &= -1 - 3t, \
z &= 5 + 3t.
\end{aligned}
$$

Thus, the parametric equations are:
$$
\begin{array}{l}
x = 2-4t, \
y = -1-3t, \
z = 5+3t.
\end{array}
$$
The parametric equations of the line are: $$ \begin{array}{l} x = 2 - 4t, \ y = -1 - 3t, \ z = 5 + 3t. \end{array} $$

Find the parametric form of the line that goes through the point
and is parallel to the vector .

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Find the parametric form of the line that goes through the point and is parallel to the vector .