just use filter

let rows = [
  [0, 'Peter', undefined],
  [3, 'John', 90909090],
  [5, 'Mary', undefined],
];

rows = rows.filter((row) => {
  return row.indexOf(undefined) === -1;
});

console.log(rows); // [ [ 3, 'John', 90909090 ] ]

if you mean that delete the array if it has a string cell contain 'undefined' as string you can go with this

let rows = [
  [0, 'Peter', 'undefined value'],
  [3, 'John', 90909090],
  [5, 'Mary', 'undefined'],
];

rows = rows.filter((row) => {
  let isContainUndefined = false;
  row.forEach((element) => {
    if (String(element).includes('undefined')) {
      isContainUndefined = true;
      return;
    }
  });
  return !isContainUndefined;
});

console.log(rows); // [ [ 3, 'John', 90909090 ] ]

With filter, then some on each item which is truthy if none of the items in each array includes undefined (string).

let rows = [
  [0, "Peter", "undefined value"],
  [3, "John", 90909090],
  [5, "Mary", "undefined"]
];

rows = rows.filter(row => !row.some(v => typeof v === 'string' && v.includes('undefined')));

console.log(rows); // [ [ 3, 'John', 90909090 ] ]