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The two prescribed directions have dot product zero. Normalize them toTheir cross product is the perpendicular unit vector ${\mathbf{q}}_3=(2,-1,-1{)}^T/\sqrt{6}$. Using these as columns givesThe columns form an orthonormal basis, so $Q^{T}Q=I$ and $Q$ is an orthogonal matrix.

Suppose $B\mathbf{x}=0$. The given quadratic form bound then givesso $\mathbf{x}=0$. Thus the kernel of $B$ is trivial. A square matrix with trivial kernel is an invertible matrix, proving the assertion. No symmetry assumption on $B$ is needed.

The cube is $1\le x\le 2$, $0\le y,z\le 1$, so $|x|=x$ throughout it. Evaluate the six outward surface integrals directly.

On $x=2$ the normal component is $2$, giving flux $2$; on $x=1$ it is $-1$, giving flux $-1$. On $y=1$ the normal component is $2x$, giving ${\int }_1^2{\int }_0^{1}2xdzdx=3$; on $y=0$ it is zero. On $z=1$ the normal component is $e^{-1}$ and on $z=0$ it is $-1$, so those unit-area faces give $e^{-1}$ and $-1$.

Adding all six contributions yieldsOnly direct surface integral was used, as requested.

The homogeneous solution is $A\cos x+B\sin x$. If $k\ne 1$, the method of undetermined coefficients permits a particular integral $C\cos kx$. Substitution gives $(1-k^2)C=1$ and henceFor $k=1$, the forcing is itself a homogeneous solution, so this trial fails by resonance in a differential equation. Multiplying the sine trial by $x$ givesThus the resonant general solution isThe secular term containing $x$ describes the increasing amplitude under undamped resonant forcing.