Given plane is \(\vec r. (2\hat i-\hat j + \hat k) + 8 = 0 \)........(1)
Required plane passes through the line of intersection of passes \(\vec r. (2\hat i-3\hat j + 4\hat k) = 1\, and \,\vec r. (\hat i-\hat j) + 4 = 0\)
∴ Equation of require plane is \(\vec r . (2\hat i-3\hat j + 4\hat k) + λ(\hat i - \hat j) + 1 + 4λ = 0\)
= \(\vec r. ((2 + λ)\hat i + (-3- λ)\hat j + 4\hat k) + 1 + 4 λ = 0 ....(2) \)
∴ Plane (1) & (2) perpendicular to each other
∴ 2 (2 + λ) + -1 x - (3 + λ) + 1 x 4 = 0
= 4 + 2λ + 3 + λ + 4 = 0
= 3λ + 11 = 0
= λ = -11/3
∴ Equation of required plane is \(\vec r. ((2-\frac {11}{3})\hat i + (-3 + \frac {11}{3}) \hat j + 4\hat k) + 1 + 4 x - \frac {11}{3} = 0\)
= \(\vec r.(-5 \hat i + 2\hat j + 12 \hat k) + 3-44 = 0\)
= \(\vec r.(-5\hat i + 2\hat j + 12 \hat k) - 41 = 0 ......(3)\)
∴ Direction ratios of normal to plane (3) are -5, 2 &12.
Given line is x-1 = 2y-4 = 3 z -12
= \(\frac {x-1}{1} = \frac {y-2}{\frac 12} = \frac {z-4}{\frac 13}\)
∴ Direction ratios of line are 1, 1/2 & 1/3 and it passes through point (1, 2, 4)
∴ -5 x 1 + 2 x 1/2 + 12 x 1/3 = -5 + 1+ 4
= -5 + 5 =0
Hence, line is perpendicular to normal of the plane (3).
∴ Line is paralleled to plane (3)
Also \((\hat i + 2\hat j + 4\hat k).(-5 \hat + 2\hat j + 12\hat k) -41 = 0\) (from (3) )
= -5 + 4 + 48 - 41 = 0
= 6 = 0 (Not satisfied)
Hence, point (1, 2, 4) does not lie on plane (3).
Therefore, given line is paralleled to plane (3) but plane did not contains the line.