If the distance between the plane Ax - y + z = d and the plane containing the lines x/2 = y/3 = z/4 and x/3 = y/4 = z/5 is √6, then the value of |d| is

From the given equations, we get
2l + 3m + 4n = 0
3l + 4m + 5n = 0
l/1 = m/2 = n/3
Equation of plane will be:
→ a(x/1) + b(y/2) + c(z/3) = 0
→ 1(x/1) + 2(y/2) + 3(z/3) = 0
→ x + y + z = 0
→ -x + 2y - z + 3 = 0
→ -x + 2y - z = -3
→ x - 2y + z = 3
The distance between the planes Ax - y + z = d and x - 2y + z = 3 is given by:
√6 = |d - 3|/√(1² + (-2)² + 1²) = |d - 3|/√6
|d - 3| = 6
d - 3 = ±6
d = 9 or d = -3
Therefore, |d| = 3 or |d| = 9