Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{\mathrm{x} + 3}{2} = \frac{\mathrm{y} - 3}{7} = \frac{\mathrm{z} - 2}{5}$

Equation of the plane passing through the point (3, 4, 1) is:

a ( x - 3 ) + b ( y - 4 ) c ( z - 1 ) = 0        .............(1)

Where a, b, c are the direction ratios of the normal to the plane 

It is given that the plane (1) passes through the point 9 0, 1, 0 ).

$\therefore$ a - 3 + b - 3 + c - 1 = 0

$\Rightarrow$3a + 3b + c = 0                                  ...............(2)

It is also given that the plane (1) is parallel to the line 

$\frac{\mathrm{x} + 3}{2} = \frac{\mathrm{y} - 3}{7} = \frac{\mathrm{z} - 2}{5}$.

So, this line is perpendicular to the normal of the plane (1).

$\therefore$ 2a + 7b + 5c = 0                                 ................(3)

Solving equations (2) and (3), we have:

$$\begin{gathered} \frac{\mathrm{a}}{\left(3 \mathrm{x}5\right) - \left(7 \mathrm{x} 1\right)} = \frac{\mathrm{b}}{\left(1 \mathrm{x} 2\right){\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \left(5 \mathrm{x} 3\right)}} = \frac{\mathrm{c}}{\left(3 \mathrm{x} 7\right){\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \left(2 \mathrm{x} 3\right)}} \\ \Rightarrow \frac{\mathrm{a}}{8} = \frac{\mathrm{b}}{-13} = \frac{\mathrm{c}}{15} \end{gathered}$$

So, the direction ratios of the normal to the required plane are multiples of 8, -13, 15.

Therefore, equation (1) becomes:

8 ( x - 3 ) -13 ( y - 4 ) + 15 ( z - 1) = 0

$\Rightarrow$ 8x - 13y + 15z + 13 = 0.

Which is the required equation of the plane.