Let direction ratios of normal of required plane are A, B and C

Since, required plane is perpendicular to the plane 

x - 2y + 4z = 10

∴ Their normal are perpendicular to each other

∴ A -2B + 4C = 0 ...(i)

Since required plane is passing through the points 

A(2, 1, -1) & B (-1, 3, 4)

∴ Normal of required plane is perpendicular to line passing through points A (2, 1, -1) & B (-1, 3, 4). 

And direction ratios of line AB are 2-(-1), 1-3, -1-4 or 3, -2, -5

∴ 3A-2B -5C = 0...(ii)

Now from (i) and (ii)

\(\frac {A}{-2\times-5 + 2 \times4} = \frac {B}{4\times3+5\times1}= \frac{C}{1\times-2-3\times-2}\)(By cross multiplication method)

= A/18 = B/17 = C/4 =  λ  Lets

∴ A = 18λ , B 17λ  & C = 4λ .

Hence, direction ratios of required plane are 18λ , 17λ  & C4λ and required plane is passing through point A(1, 1,-1).

∴ Equation of required plane is 

A(x-2) + B (y-1) + C (z+1) = 0

= 18λ (x-2) + 17λ (y-1) + 4λ (z+1) = 0

= 18(x-2) + 17(y-1) + 4(z+1) = 0

= 18x + 17y+4z - 36-17+4=0

= 18x + 17y + yz -49 = 0...(iii)

Given line \(\oversetλ r\) = (\((\hat-i +3\hat j + 4\hat k) +λ (3\hat i-2\hat j-5\hat k)\)

∴ direction vector of line \(\oversetλ r\) is \(\hat b = 3\hat i- 2\hat j-5\hat k\) and line \(\oversetλ r\) is passing through point whose position vector is \(\oversetλ a\) = \(-\hat i + 3\hat j + 4\hat k\)

(By comparing \(\oversetλ r\) = \(\oversetλ a\) + λ \(\oversetλ b\))

Thus, the direction ratios of line \(\oversetλ r\) = are 3, -2, -5 & line \(\oversetλ r\) is direction vector of line \(\oversetλ r\) is perpendicular to normal of plane (3) and point (-1, 3, 4) lies on the plane (3)

Since, 18 x 3 + 17 x -2 + 4 x -5 = 54 -34 -20

= 54 - 54 = 0

It implies that direction vector of line \(\oversetλ r\) is perpendicular to the normal of the plane.

Also, 18 x -1 + 17 x 3+ 4 x 4 -49 = -18 + 51 + 16 - 49 = 67 - 67 = 0

It implies that point  (-1, 3, 4) lies on the plane (3) we conclude that plane 18 x + 17 y + 4z - 49 = 0

contains the line \(\oversetλ r\) = \((-\hat i+ 3\hat j + 4\hat k) + λ (3\hat i-2 \hat j- 5\hat k)\)

Hence Proved.

Plane passes through (2,1,-1) and (-1,3,4), 

A(x - 2) + B(y -1) + C(z + 1) = 0 (1) 

A(x + 1) + B(y - 3) + C(z - 4) = 0 (2) 

Subtracting (1) from (2), 

A(x + 1 - x + 2) + B(y - 3 - y + 1) + C(z - 4 - z -1) = 0 

3A - 2B - 5C = 0 (3) 

Now plane is perpendicular to x - 2y + 4z = 10 

A - 2B + 4C = 0 (4) 

Using (3) in (4) 

2A - 9C = 0

C = \(\frac{2}9\)A

2B = A + 4. \(\frac{2}9\)A

⇒ \((\frac{9+8}9)\)A = \(\frac{17}9\)A

B = \(\frac{17}{18}\)A

Putting values in equation (1)

A(x - 2) + \(\frac{17}{18}\)A (y - 1) + \(\frac{2}9\)A (z + 1) = 0

A(18(x-2) + 17(y-1) + 4(z + 1) = 0 

18x + 17y + 4z - 36 -17 + 4 = 0 

18x + 17y + 4z - 49 = 0 

So, the required equation of plane is 18x + 17y + 4z - 49 = 0

If plane contains  \(\bar{r}\) = -\(\hat{i}\) + 3\(\hat{j}\) + 4\(\hat{k}\) + (3\(\hat{i}\) - 2\(\hat{j}\) - 5\(\hat{k}\)) then (-1,3,4) satisfies plane and normal vector of plane is perpendicular to vector of line

LHS =18(-1) + 17.3 + 4.4 - 49 

= -18 + 51 + 16 - 49 

= -2 + 2 = 0 = RHS 

In vector form normal of plane 

\(\bar{n}\) = 18\(\hat{i}\) + 17\(\hat{j}\) + 4\(\hat{k}\)

LHS =18.3 + 17(-2) + 4.(- 5) = 54 - 34 - 20 = 0 = RHS 

Hence line is contained in plane.