If [tex]\hat\gamma-c\hat\alpha-d\hat\beta[/tex] is perpendicular to both [tex]\hat\alpha[/tex] and [tex]\hat\beta[/tex], then their dot products are zero.
Recall that for any vector [tex]\hat x[/tex], [tex]\hat x\cdot \hat x = \|\hat x\|^2[/tex].
Now, we have
[tex](\hat\gamma - c\hat\alpha - d\hat\beta) \cdot \hat\alpha = (\hat\gamma\cdot\hat\alpha) - c\|\hat\alpha\|^2 - d (\hat\beta\cdot\hat\alpha) = 0[/tex]
[tex]\hat\gamma\cdot\hat\alpha = 5 + 0 - 16 = -11[/tex]
[tex]\|\hat\alpha\|^2 = 1^2 + 1^2 + (-2)^2 = 6[/tex]
[tex]\hat\beta\cdot\hat\alpha = -1 + 2 - 6 = -5[/tex]
[tex]\implies 6c - 5d = -11[/tex]
and
[tex](\hat\gamma - c\hat\alpha - d\hat\beta) \cdot \hat\beta = (\hat\gamma\cdot\hat\beta) - c(\hat\alpha\cdot\hat\beta) - d \|\hat\beta\|^2 = 0[/tex]
[tex]\hat\gamma\cdot\hat\beta = -5 + 0 + 24 = 19[/tex]
[tex]\|\hat\beta\|^2 = (-1)^2 + 2^2 + 3^2 = 14[/tex]
[tex]\implies 5c - 14d = -19[/tex]
Solve for [tex]c[/tex] and [tex]d[/tex]. By elimination,
[tex]5(6c - 5d) - 6(5c - 14d) = 5(-11) - 6(-19)[/tex]
[tex]-25d + 84d = -55 + 104[/tex]
[tex]59d = 59[/tex]
[tex]\boxed{d=1}[/tex]
[tex]6c-5 = -11[/tex]
[tex]6c=-6[/tex]
[tex]\boxed{c=-1}[/tex]