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The polar form is $z=e^{3\pi i/4}$. Interpreting the fractional complex exponentiation as all its values, a candidate $w$ satisfies $w^3=z^2=-i$. The three roots of a complex number areThey have arguments $\pi /2$, $7\pi /6$ and $11\pi /6$, respectively. Since all have modulus one, their squared distances from $z$ in the complex plane are $|w-z{|}^2=2-2\cos (\arg w-3\pi /4)$, givingThe last is strictly largest, so the required root of a complex number isRestricting complex exponentiation to its principal value would give only $i$; the request to choose among roots requires all three values.

Apply the vector triple product identity $p\times (c\times d)=c(p\cdot d)-d(p\cdot c)$ with $p=a\times b$. This givesFor column vectors $a,b,d$, the scalar triple product is $(a\times b)\cdot d=\det (a,b,d)$; replacing $d$ by $c$ gives the other coefficient. ThereforeThe signs follow the stated column convention and the cyclic invariance of the scalar triple product. No independence assumptions on the four vectors are needed.

Write the given matrix as $A$. Expanding its characteristic polynomial along the middle column givesThus the eigenvalues are $1,-2,-2$, counted with algebraic multiplicity.

For $\lambda =1$, solving $(A-I)v=0$ gives $z=2x$ and $y=-x$. For $\lambda =-2$, solving $(A+2I)v=0$ gives $z=3x$, with $y$ free. A convenient eigenbasis is thereforeThe last two eigenvectors span the two-dimensional eigenspace for $-2$. The three columns have determinant $-1$, so they are linearly independent and form a basis of ${\mathbb{R}}^3$. In particular the repeated eigenvalue causes no failure of diagonalizability here.

Let $Av=\lambda v$ with $v\ne 0$, using the standard complex Hermitian inner product. ThenTaking the complex conjugate and using $A^{†}=-A$ yields $\overline{\lambda }{v}^{†}v=v^{†}{A}^{†}v=-\lambda v^{†}v$. Since $v^{†}v>0$,Thus every eigenvalue of an anti-Hermitian matrix lies on the imaginary axis, including the possible zero eigenvalue; each nonzero eigenvalue is purely imaginary.

Now take $Au=\lambda u$ and $Av=\mu v$. On one hand $u^{†}Av=\mu u^{†}v$. On the other handConsequently $(\mu -\lambda )u^{†}v=0$. Distinct eigenvalues therefore implywhich is the required orthogonality of the eigenvectors.

A normal vector to the plane is $n=(3,4,12{)}^T$, with Euclidean norm $\Vert n\Vert =\sqrt{9+16+144}=13$. For $p=(1,1,1{)}^T$, $n\cdot p=19$, so the distance from a point to a plane isTo verify that this is the minimum, the displacement to the closest point on a plane is parallel to $n$. Put $q=p+tn$ and require $n\cdot q=26$. This gives $t=7/169$ andFor every other point $r$ on the plane, the displacement $r-q$ is orthogonal to $q-p$. The Pythagorean theorem gives $\Vert r-p{\Vert }^2=\Vert q-p{\Vert }^2+\Vert r-q{\Vert }^2\ge (7/13{)}^2$, proving minimality.

On the interval $x>0$ containing the initial point, divide the linear ordinary differential equation by $x$ to obtain $y^{\prime }-2y/x=x$. Its integrating factor is $e^{\int -2dx/x}=x^{-2}$, givingThe initial condition gives $1=C$. HenceDirect differentiation gives $y^{\prime }=x(3+2\ln x)$, so $x{y}^{\prime }-2y=x^2$ and $y(1)=1$. The coefficient of $y^{\prime }$ vanishes at zero; the boxed expression is the solution on the regular initial-value interval, rather than an assertion of unique continuation through zero.

The right-hand side depends on $x+y$, suggesting the substitution $u=x+y$. Then $y^{\prime }=u^{\prime }-1$, so the ordinary differential equation becomesThe original equation is undefined at $u=0$. If $u\ne 2$, this is a separable differential equation:Substituting back gives the implicit familyDifferentiating the implicit equation gives $y^{\prime }-1+2(1+y^{\prime })/(x+y-2)=0$, or $(x+y)y^{\prime }=x+y-4$, verifying it on its permitted intervals.

Division by $u-2$ excluded the constant solution $u=2$, which must be restored separately:It satisfies $y^{\prime }=-1=(2-4)/2$. Together these give all local real solutions on intervals where $x+y\ne 0$.