3D..........

We can think of the required line as the intersection of two planes \sigma_3=0 and \sigma_4=0 which are as follows:
i) The plane \sigma_3=0 passes through (4,-14,4) and is perpendicular to the two given planes
\sigma_1 = 3x+2y-z-5=0 and \sigma_2 = x-2y-2z+1=0

ii) The plane \sigma_4=0 passes through (4,-14,4) and the intersection of the planes \sigma_1=0, \ \sigma_2=0

The plane \sigma_3 is of the form
a_1(x-4)+b_1(y+14)+c_1(z-4)=0
Since it is perpendicular to both \sigma_1=0, \ \sigma_2=0, we have
3a₁ + 2b₁ - c₁=0 and
a₁ - 2b₁ - 2c₁ = 0
From where, we obtain, by cross multiplication,
\dfrac{a_1}{-6}=\dfrac{b_1}{5}=\dfrac{c_1}{-8}
Hence, the plane \sigma_3 is

-6(x-4) + 5(y+14)-8(z-4)=0

i.e 6 x - 5y + 8 z =126

And the plane \sigma_4 will be a linear combination of \sigma_1 and \sigma_2
i.e. \sigma_4 will be of the form

3x+2y-z - 5 + λ (x-2y-2z +1)=0

Since this plane passes through (4,-14,4), we get
3(4) + 2(-14) - 4 - 5+ λ(4 - 2(-14)-2(4) +1)=0

i.e λ = 1.

So \sigma_4 is

4x -3z -4 =0

So the required line is given by

6 x - 5y + 8 z -126 = 0 = 4x -3z -4