If $(x, y, z) ≠ (0, 0, 0)$ and $(\hat i+\hat j+3\hat k)x+(3\hat i-3\hat j+\hat k) y+(-4\hat i+5\hat j) z =a (x\hat i + y\hat j +z\hat k)$,  then the values of a are

0, -1

The given vector equation can be written as

$[(1-a) x + 3y - 4z]\hat i + [x-(3+ a) y +5 z]\hat j + [3x + y −az] \hat k=\vec 0$

∵ Since $\hat i,\hat j,\hat k$ are linearly independent vectors.

$∴(1-a) x + 3y-4z=0$

$x-(3+a) y +5z = 0$

$3x+y-az=0$

Eliminating $x, y, z$, we get

$\begin{vmatrix}1-a&3&- 4\\1&-(3+ a)&5\\3&1&-a\end{vmatrix}=0⇒a=0,-1$