To be honest I don't completely understand this problem, by definition - cross product is a vector and this vector is orthogonal to vectors involving in cross product, but here i see that vector of cross product is a linear combination of vectors $a$ and $x$, so it's means that this vector is coplanar to $a$ and $x$, am I wrong?
But okay, I did this following
Let $$a\times x=b$$ Then $$x =\alpha a +\beta[a\times b]$$ Where $\alpha$ and $\beta$ are projections, hence $$\alpha=(x\cdot a)/|a|^2$$
$$\beta=1/|a|^2$$ $$[a\times b]=[a\times(3/4x-1/4a)]$$ $$[a\times b]=3/4[a\times b]$$ $$x =\alpha a + 3/4\beta(1/4a-3/4x)$$ it's leaves us with $$x = (16\alpha+3\beta)/(16+9\beta)a$$ so I've got that this vectors are colinear, it's very confusing, $$x=(16(x\cdot a)+3)/(16|a|^2+9)a$$ no matter what I'm doing I'm still having this result, but you all see that in this case $x$ and $a$ are zero vectors, idk, where is my mistake?
You are right that normally the cross product of two vectors cannot be co-planar to those vectors, unless something turns out to be zero - either $a$, $b$, or $a\times b$. If $a$ or $b$ is zero, so is $a\times b$, and the equation forces both $a$ and $b$ to be zero. If $a\times b$ is zero, then the equation becomes $$0=\frac{3}{4}x - \frac{1}{4}a$$ with the solution $$a=3x .$$ That is, the two vectors are parallel - another situation in which the cross product is zero.