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(i+j+3k)x+(3i-3j+k)y+(-4i+5j)z=lambda(x ...

`(i+j+3k)x+(3i-3j+k)y+(-4i+5j)z=lambda(x i+yj+zk)` then lambda equal to

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`(hati+hatj+3hatk)x+(3hati-3hatj+hatk)y +(-4hati+5hatj)z = lambda(xhati+yhatj+zhatk)`
Comparing coefficient of `hati, hatj and hatk`
`x+3y-4z = lambda x =>(1-lambda)x+3y-4z = 0->(1)`
`x-3y+5z = lambda y => x+(-3-lambda)y+5z = 0->(2)`
`3x+y = lambda z => 3x+y-lambdaz = 0->(3)`
Now, solving (1),(2),(3) using determinant,
`|[1-lambda,3,-4],[1,-3-lambda,5],[3,1,-lambda]| = 0`
`=>[(1-lambda)(3lambda+lambda^2-5)-3(-lambda-15)-4(1+9+3lambda)] = 0`
...
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