How to find the equation of the plane through points $(2, -1, 0)$ and $ (3, 4, 5)$ and parallel to the line $2x=3y=4z$.
If i find out the direction ratios: $$a : b : c = (3-2) : (4+1) : (5-0) = 1 : 5 : 5$$
As we know general form of a plane is $ax + by + cy + d = 0$
Now how to go forward?
We use following formula which is for equation of a plane passing through a given straight line($l_1)$ and parallel to another given line $(l_2)$:
${\begin{vmatrix}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}}=0$
Let line $l_1$ passing through the given points (2, -1, 0) and (3, 4, 5) be:
$\frac {x-2}1=\frac{y+1}5=\frac z5\Rightarrow l_1=1, m_1=5, n_1=5$
and line $l_2$ be:
$2x=3y=4z\rightarrow \frac x{\frac12}=\frac y{\frac13}=\frac z{\frac14}\Rightarrow l_2=\frac 12, m_2=\frac 13, n_2=\frac 14$
Now plugging values in formula we obtain:
$(x-2)\begin{vmatrix}5&5\\\frac13&\frac14\end{vmatrix}+(y+1)\begin{vmatrix}5&1\\\frac14&\frac12\end{vmatrix}+(z)\begin{vmatrix}1&5\\\frac12&\frac13\end{vmatrix}=0$
Now calculate determinants and find the required equation.