Q.Find the Cartesian equations of the line which passes through the point $(-2, 4, -5)$ and parallel to the line $\dfrac{x+2}{3} = \dfrac{y-3}{5} = \dfrac{z+5}{6}$.

Apply the standard cartesian and vector line-equation formulas for a point and direction ratios.

Point $A(1,2,3)$, direction ratios $2,3,7$.

Cartesian equation:

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{7}$

Vector equation: with $\bar a=\hat i+2\hat j+3\hat k$ and $\bar b=2\hat i+3\hat j+7\hat k$,

$\bar r=\bar a+\lambda\bar b=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+7\hat k)$

The required line's direction is the cross product of the two given lines' direction vectors.

Direction ratios of the two given lines: $\vec{u}=(1,2,3)$ and $\vec{v}=(-3,2,5)$.

$\vec{u}\times\vec{v} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & 2 & 3\\ -3 & 2 & 5\end{vmatrix} = \hat i(10-6)-\hat j(5+9)+\hat k(2+6) = (4,-14,8)$

Dividing by 2: direction ratios $(2,-7,4)$.

Line through $(3,1,2)$ with these direction ratios:

$\dfrac{x-3}{2}=\dfrac{y-1}{-7}=\dfrac{z-2}{4}$

Rewrite each expression as (variable $-$ constant)/(denominator) by introducing a common parameter.

$2x-2 = 3y+1 = 6z-2$. Let this common value be $t$.

From $2x-2=t$: $x = 1+\dfrac t2$

From $3y+1=t$: $y = -\dfrac13+\dfrac t3$

From $6z-2=t$: $z = \dfrac13+\dfrac t6$

So, parametrically: $x=1+\dfrac t2,\ y=-\dfrac13+\dfrac t3,\ z=\dfrac13+\dfrac t6$, giving direction ratios $\dfrac12,\dfrac13,\dfrac16$, i.e. (multiplying by 6) $3, 2, 1$, through the point $\left(1,\,-\dfrac13,\,\dfrac13\right)$.

Cartesian (symmetric) form:

$\dfrac{x-1}{3} = \dfrac{y+\frac13}{2} = \dfrac{z-\frac13}{1}$

Vector equation:

$\vec r = \left(\hat i-\dfrac13\hat j+\dfrac13\hat k\right) + \lambda\left(3\hat i+2\hat j+\hat k\right)$