Find the equation of the plane through the points (2,1,-1),(-1,3,4) and perpendicular to the plane $x - 2y + 4z = 10$.

The equation of the plane passing through (2,1,-1) is $a(x - 2) + b(y - 1) + c(z + 1) = 0$

……..(i)
Sicne, this passes through (-1,3,4) $\therefore a( - 1 - 2) + b(3 - 1) + c(4 + 1) = 0$

$\Rightarrow$ $- 3a + 2b + 5c = 0$

……..(ii)
Since, the plane (i) is perpendicular to the plane $x - 2y + 4z = 10$. $\therefore 1 \cdot a - 2 \cdot b + 4 \cdot c = 0$
$\Rightarrow$ $a - 2b + 4c = 0$

……(iii)
On solving Eqs. (ii) and (iii),

we get
$\frac{a}{{8 + 10}} = \frac{{ - b}}{{ - 17}} = \frac{c}{4} = \lambda$

$\Rightarrow$ $a = 18\lambda ,b = 17\lambda ,c = 4\lambda$

From Eq. (i),
$18\lambda (x - 2) + 17\lambda (y - 1) + 4\lambda (z + 1) = 0$

$\Rightarrow$ $18x - 36 + 17y - 17 + 4z + 4 = 0$

$\Rightarrow$ $18x + 17y + 4z - 49 = 0$
$\therefore 18x + 17y + 4z = 49$