Step 1: Use the condition that \( \mathbf{e} \) is perpendicular to \( \mathbf{i} + \mathbf{j} + \mathbf{k} \).
Since \( \mathbf{e} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is perpendicular to the vector \( \mathbf{i} + \mathbf{j} + \mathbf{k} \), we have the dot product condition:
\[
\mathbf{e} \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0
\]
\[
a + b + c = 0
\]
Step 2: Use the condition that \( \mathbf{e} \) is coplanar with \( \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \) and \( 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} \).
For coplanarity, the scalar triple product of \( \mathbf{e}, \mathbf{i} - 3\mathbf{j} + 5\mathbf{k}, \) and \( 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} \) must be zero:
\[
\begin{vmatrix}
a & b & c
1 & -3 & 5
3 & 1 & -5
\end{vmatrix} = 0
\]
Expanding the determinant:
\[
a \left( (-3)(-5) - (5)(1) \right) - b \left( (1)(-5) - (5)(3) \right) + c \left( (1)(1) - (-3)(3) \right) = 0
\]
\[
a(15 - 5) - b(-5 - 15) + c(1 + 9) = 0
\]
\[
a(10) - b(-20) + c(10) = 0
\]
\[
10a + 20b + 10c = 0
\]
Step 3: Solve the system of equations.
We now have the system of equations:
1. \( a + b + c = 0 \)
2. \( 10a + 20b + 10c = 0 \)
From the first equation, solve for \( c \):
\[
c = -a - b
\]
Substitute into the second equation:
\[
10a + 20b + 10(-a - b) = 0
\]
\[
10a + 20b - 10a - 10b = 0
\]
\[
10b = 0
\]
\[
b = 0
\]
Substitute \( b = 0 \) into \( a + b + c = 0 \):
\[
a + c = 0 \quad \Rightarrow \quad c = -a
\]
Step 4: Use the unit vector condition.
Since \( \mathbf{e} \) is a unit vector:
\[
a^2 + b^2 + c^2 = 1
\]
Substitute \( b = 0 \) and \( c = -a \):
\[
a^2 + 0^2 + (-a)^2 = 1
\]
\[
2a^2 = 1 \quad \Rightarrow \quad a^2 = \frac{1}{2}
\]
\[
a = \pm \frac{1}{\sqrt{2}}
\]
Now, using \( c = -a \), we find:
\[
c = \mp \frac{1}{\sqrt{2}}
\]
Step 5: Compute \( 2a^2 + 3b^2 + 4c^2 \).
Substitute \( a^2 = \frac{1}{2} \), \( b = 0 \), and \( c^2 = \frac{1}{2} \) into the expression \( 2a^2 + 3b^2 + 4c^2 \):
\[
2a^2 + 3b^2 + 4c^2 = 2 \times \frac{1}{2} + 3 \times 0 + 4 \times \frac{1}{2}
\]
\[
= 1 + 0 + 2 = 3
\]
Thus, the correct value of \( 2a^2 + 3b^2 + 4c^2 \) is \( \boxed{3} \).
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