연립미분방정식 { x ′ ( t ) = 7 x ( t ) − y ( t ) + 6 z ( t ) y ′ ( t ) = − 10 x ( t ) + 4 y ( t ) − 12 z ( t ) z ′ ( t ) = − 2 x ( t ) + y ( t ) − z ( t ) \begin{cases} x' \left (t \right) =7x \left (t \right) -y \left (t \right) +6z \left (t \right) \\ y' \left (t \right) = -10x \left (t \right) +4y \left (t \right) -12z \left (t \right) \\ z' \left (t \right) = -2x \left (t \right) +y \left (t \right) -z \left (t \right) \end{cases} ⎩ ⎨ ⎧ x ′ ( t ) = 7 x ( t ) − y ( t ) + 6 z ( t ) y ′ ( t ) = − 10 x ( t ) + 4 y ( t ) − 12 z ( t ) z ′ ( t ) = − 2 x ( t ) + y ( t ) − z ( t ) 의 해 ( x ( t ) y ( t ) z ( t ) ) \begin{pmatrix*}[r] x \left (t \right) \\ y \left (t \right) \\ z \left (t \right) \end{pmatrix*} x ( t ) y ( t ) z ( t ) 가 초기조건 ( x ( 0 ) y ( 0 ) z ( 0 ) ) = ( − 1 4 2 ) \begin{pmatrix*}[r] x \left (0 \right) \\ y \left (0 \right) \\ z \left (0 \right) \end{pmatrix*} = \begin{pmatrix*}[r] -1 \\ 4 \\ 2 \end{pmatrix*} x ( 0 ) y ( 0 ) z ( 0 ) = − 1 4 2 를 만족할 때, x ( 1 ) + y ( 1 ) + z ( 1 ) x \left (1 \right) +y \left (1 \right) +z \left (1 \right) x ( 1 ) + y ( 1 ) + z ( 1 ) 의 값은 a e l + b e m + c e n a e ^ {l} +b e ^ {m} +c e ^ {n} a e l + b e m + c e n 이다. a + b + c + l + m + n a+b+c+l+m+n a + b + c + l + m + n 의 값을 구하시오.
(단, a , b , c , l , m , n a, b, c, l, m, n a , b , c , l , m , n 은 모두 정수이다.)