Expand each vector as linear combinations of the standard basis vectors:
(7, 9, 6) = 7(1, 0, 0) + 9(0, 1, 0) + 6(0, 0, 1)
(-4, 1, 5) = -4(1, 0, 0) + (0, 1, 0) + 5(0, 0, 1)
For brevity, write
i = (1, 0, 0)
j = (0, 1, 0)
k = (0, 0, 1)
Then by definition of the cross product,
i x i = j x j = k x k = (0, 0, 0)
i x j = k
j x k = i
k x i = j
and for any two vectors a and b, we have a x b = - b x a.
Now compute the product:
(7i + 9j + 6k) x (-4i + j + 5k)
= -28 (i x i) - 36 (j x i) - 24 (k x i)
... + 7 (i x j) + 9 (j x j) + 6 (k x j)
... + 35 (i x k) + 45 (j x k) + 30 (k x k)
= -36 (-k) - 24 j + 7 k + 6 (-i) + 35 (-j) + 45 i
= 39 i - 59 j + 43 k
which is the same as the vector
(39, -59, 43)
And yes, this vector is perpendicular to both given vectors.