Solutions:
Let projection of \( L_1 \) on plane \( P \) is line \( L_2 \).
\( P: x - 2y - z - 3 = 0 \)
\( L_1 : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2} = \lambda \)
'\( A \)' is the point of intersection of \( L_1 \) & \( P \).
\( A(2\lambda + 1,\ \lambda + 3,\ 2\lambda + 4) \) satisfies equation of \( P \).
\( \Rightarrow (2\lambda + 1) - 2(\lambda + 3) - (2\lambda + 4) - 3 = 0 \)
\( \Rightarrow \lambda = -6 \Rightarrow A(-11,\ -3,\ -8) \)
Since, \( L_1 \) passes through \( C(1,3,4) \). \( B \) is the projection of \( C \) in \( P \).
Also, \( B \) is the foot of perpendicular from \( C \) on \( P \)
\( \Rightarrow \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \)
\( \Rightarrow \frac{x_2 - 1}{1} = \frac{y_2 - 3}{-2} = \frac{z_2 - 4}{-1} = \frac{(1 - 2(3) - (4) - 3)}{1 + 4 + 1} \)
\( \Rightarrow B(x_2, y_2, z_2) = B(3,\ -1,\ 2) \)
Now, \( L_2 \) passes from \( A \) & \( B \)
\( \Rightarrow L_2 : \frac{x - 3}{14} = \frac{y + 1}{2} = \frac{z - 2}{10} \ (\text{OR})\ \frac{x - 3}{7} = \frac{y + 1}{1} = \frac{z - 2}{5} = \mu \)
Let \( E(7\mu + 3,\ \mu - 1,\ 5\mu + 2) \) lies on \( L_2 \) such that \( DE \perp L_2 \) where \( D(0,0,6) \)
\( \Rightarrow \) D.R.s of \( DE \) = \( \langle 7\mu + 3,\ \mu - 1,\ 5\mu - 4 \rangle \) & D.R.s of \( L_2 \) = \( \langle 7,\ 1,\ 5 \rangle \)
\( DE \perp L_2 \Rightarrow 7(\mu + 3) + 1(\mu - 1) + 5(5\mu - 4) = 0 \Rightarrow \mu = 0 \Rightarrow E(3,\ -1,\ 2) \)
Now, \( DE = \sqrt{9 + 1 + 16} = d \Rightarrow d^2 = 26 \)

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